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Page No 32.14:

Question 1:

Which of the following distributions of probabilities of a random variable X are the probability distributions?
(i)

X : 3 2 1 0 −1
P (X) : 0.3 0.2 0.4 0.1 0.05

(ii)
X : 0 1 2
P (X) : 0.6 0.4 0.2

(iii)
X : 0 1 2 3 4
P (X) : 0.1 0.5 0.2 0.1 0.1

(iv)
X : 0 1 2 3
P (X) : 0.3 0.2 0.4 0.1

Answer:

(i) P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) + P (X = -1)
    = 0.3 + 0.2 + 0.4 + 0.1 + 0.05
    =1.05 > 1
It is not the probability distribution of random variable X.

(ii) P (X = 0) + P (X = 1) + P (X = 2)
     = 0.6 + 0.4 + 0.2
     = 1.2 > 1
It is not the probability distribution of random variable X.

(iii) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
      = 0.1 + 0.5 + 0.2 + 0.1 + 0.1
      = 1
It is the probability distribution of random variable X.

(iv) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
      = 0.3 + 0.2 + 0.4 + 0.1
      = 1
It is the probability distribution of random variable X.

Page No 32.14:

Question 2:

A random variable X has the following probability distribution:

Values of X : −2 −1 0 1 2 3
P (X) : 0.1 k 0.2 2k 0.3 k
Find the value of k.

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

0.1+k+0.2+2k+0.3+k=14k+0.6=1k=0.44=0.1

Page No 32.14:

Question 3:

A random variable X has the following probability distribution:

Values of X : 0 1 2 3 4 5 6 7 8
P (X) : a 3a 5a 7a 9a 11a 13a 15a 17a
Determine:
(i) The value of a
(ii) P (X < 3), P (X ≥ 3), P (0 < X < 5).

Answer:

(i) Since the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

a+3a+5a+7a+9a+11a+13a+15a+17a=181a=1a=181

(ii) P (X < 3)

 =PX=0+PX=1+PX=2=181+381+581=981=19

P (X ≥ 3)

PX=3+PX=4+PX=5+PX=6+PX=7+PX=8=781+981+1181+1381+1581+1781=7281=89

P (0 < X < 5)
 PX=1+PX=2+PX=3+PX=4=381+581+781+981=2481=827

Page No 32.14:

Question 4:

The probability distribution function of a random variable X is given by

xi : 0 1 2
pi : 3c3 4c − 10c2 5c − 1
where c > 0
Find: (i) c (ii) P (X < 2) (iii) P (1 < X ≤ 2)

Answer:

(i) We know that the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) = 1

3c3+4c-10c2+5c-1=13c3-10c2+9c-2=0c-13c2-7c+2=0c-13c-1c-2=0c= 13, 1, 2Neglecting 1 and 2 as individual probability should not be greater than one

(ii) P (X < 2) 

=PX=0+PX=1=3c3+4c-10c2=19+43-109=1+12-109=39=13

(iii) P (1 < X ≤ 2)

  =PX=2=5c-1=53-1=5-33=23

Page No 32.14:

Question 5:

Let X be a random variable which assumes values x1, x2, x3, x4 such that 2P (X = x1) = 3P (X = x2) = P (X = x3) = 5 P (X = x4). Find the probability distribution of X.

Answer:

Let P (X = x3) = k. Then,
P (X = x1) = k2

P (X = x2) = k3

P (X = x4) = k5

We know that the sum of probabilities in a probability distribution is always 1.

P (X = x1) + P (X = x2) + P (X = x3) + P (X = x4) = 1

k2+k3+k+k5=115k+10k+30k+6k30=161k30=1k=3061

Now,
 

xi pi
x1 k2 = 1561
x2 k3 = 1061
x3 k = 3061
x4 k5= 661

Page No 32.14:

Question 6:

A random variable X takes the values 0, 1, 2 and 3 such that:
P (X = 0) = P (X > 0) = P (X < 0); P (X = −3) = P (X = −2) = P (X = −1); P (X = 1) = P (X = 2) = P (X = 3). Obtain the probability distribution of X.

Answer:

Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)
P (X > 0) = k
    P (X < 0) = k

P (X = 0) + P (X > 0) + P (X < 0) = 1

k+k+k=1k=13

Now,
P (X < 0) = k

PX=-1+PX=-2+PX=-3=k3PX=-1=k                              PX=-1=PX=-2=PX=-3PX=-1=k3PX=-1=13×13=19 PX=-1=PX=-2=PX=-3=19Similarly, PX>0=kPX=1=PX=2=PX=3=19


Thus, the probability distribution is given by
 

Xi Pi
-3 19
-2 19
-1   19  
1 19
2 19
3 19

Page No 32.14:

Question 7:

Two cards are drawn from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=48C252C2=22562652=188221PX=1=P1 ace=4C1×48C152C2=1921326=32221PX=2=P2 aces=4C252C2=61326=1221

Thus, the probability distribution of X is given by
 

X P (X)
0 188221
1 32221
2 1221

Page No 32.14:

Question 8:

Find the probability distribution of the number of heads, when three coins are tossed.

Answer:

Let X denote the number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PTTT=18, PX=1=PHTT or TTH or THT=38PX=2=PHTH or THH or HHT=38, PX=3=PHHH=18

Thus, the probability distribution of X is given by
 

X P (X)
0 18
1 38
2 38
3 18

Page No 32.14:

Question 9:

Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 4 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno ace=48C452C4PX=1=P1 ace=4C1×48C352C4PX=2=P2 aces=4C2×48C252C4PX=3=P3 aces=4C3×48C152C4PX=4=P4 aces=4C452C4

Thus, the probability distribution of X is given by
 

X P(X)
0 48C452C4
1 4C1×48C352C4
2 4C2×48C252C4
3 4C3×48C152C4
4 4C452C4

Page No 32.14:

Question 10:

A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.

Answer:

Let X denote the number of red balls in a sample of 3 balls drawn from a bag containing 4 red and 6 black balls. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno red ball=6C310C3=20120=16PX=1=P1 red ball=4C1×6C210C3=60120=12PX=2=P2 red balls=4C2×6C110C3=36120=310PX=3=P3 red balls=4C310C3=4120=130

Thus, the probability distribution of X is given by
 

X P(X)
0 16
1 12
2 310
3 130

Page No 32.14:

Question 11:

Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.

Answer:

Let X denote the number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective mango=15C420C4=13654845=91323PX=1=P1 defective mango=5C1×15C320C4=22754845=455969PX=2=P2 defective mangoes=5C2×15C220C4=10504845=70323PX=2=P3 defective mangoes=5C3×15C120C4=1504845=10323PX=3=P4 defective mangoes=5C420C4=54845=1969

Thus, the probability distribution of X is given by
 

X P(X)
0 91323
1 455969
2 70323
3 10323
4 1969



Page No 32.15:

Question 12:

Two dice are thrown together and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?

Answer:

Let X denote the sum of the numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Sample space : {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
                            (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
                            (3, 1), (3, 2), (3 ,3), (3, 4), (3, 5), (3, 6)
                            (4, 1), (4, 2), (4 ,3), (4, 4), (4, 5), (4, 6)
                            (5, 1), (5, 2), (5 ,3), (5, 4), (5, 5), (5, 6)
                            (6, 1), (6, 2), (6 ,3), (6, 4), (6, 5), (6, 6)}

Now,
PX=2=136PX=3=236PX=4=336PX=5=436PX=6=536PX=7=636PX=8=536PX=9=436PX=10=336PX=11=236PX=12=136

Thus, the probability distribution of X is given by
 

X P(X)
2 136
3 236
4 336
5 436
6 536
7 636
8 536
9 436
10 336
11 236
12 136

Page No 32.15:

Question 13:

A class has 15 students whose ages are 14, 17, 15, 14, 21, 19, 20, 16, 18, 17, 20, 17, 16, 19 and 20 years respectively. One student is selected in such a manner that each has the same chance of being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X?

Answer:

Here, X can take the values 14, 15, 16, 17, 19, 20 and 21.
Now,
PX=14=215PX=15=115PX=16=215PX=17=315PX=18=115PX=19=215PX=20=315PX=21=115

Thus, the probability distribution of X is given by
 

X P(X)
14 215
15 115
16 215
17 315
18 115
19 215
20 315
21 115

Page No 32.15:

Question 14:

Five defective bolts are accidently mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer:

Let X denote the number of defective bolts in a sample of 4 bolts drawn from a bag containing 5 defective bolts and 20 good bolts. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective bolts=20C425C4=484512650=9692530PX=1=P1 defective bolt=5C1×20C325C4=570012650=114253PX=2=P2 defective bolts=5C2×20C225C4=190012650=38253PX=3=P3 defective bolts=5C3×20C125C4=20012650=4253PX=4=P4 defective bolts=5C425C4=512650=12530

Thus, the probability distribution of X is given by
 

X P(X)
0 9692530
1 114253
2 38253
3 4253
4 12530

Page No 32.15:

Question 15:

Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4852=12×1213×13=144169PX=1=P1 ace=452×4852=2×1213×13=24169PX=2=P2 aces=452×452=1×113×13=1169

Thus, the probability distribution of X is given by
 

X P(X)
0 144169
1 24169
2 1169

Page No 32.15:

Question 16:

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of kings.

Answer:

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno kings=4852×4852=12×1213×13=144169PX=1=P1 king=452×4852=2×1213×13=24169PX=2=P2 kings=452×452=1×113×13=1169

Thus, the probability distribution of X is given by
 

X P(X)
0 144169
1 24169
2 1169

Page No 32.15:

Question 17:

Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4751=22562652=188221PX=1=P1 ace=452×4851+4852×451=3842652=32221PX=2=P2 aces=452×351=122652=1221

Thus, the probability distribution of X is given by
 

X P(X)
0 188221
1 32221
2 1221

Page No 32.15:

Question 18:

Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Answer:

Let X denote the number of white balls in a sample of 3 balls drawn from a bag containing 4 white and 6 red balls. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno white ball=6C310C3=20120=16PX=1=P1 white ball=4C1×6C210C3=60120=12PX=2=P2 white balls=4C2×6C110C3=36120=310PX=3=P3 white balls=4C310C3=4120=130

Thus, the probability distribution of X is given by
 

X P(X)
0 16
1 12
2 310
3 130

Page No 32.15:

Question 19:

Find the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Answer:

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.

When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combinations:

(3, 6) (4, 5) (5, 4) (6, 3)

So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.

Probability of getting a total of 9 = 436=19

Probability of not getting a total of 9 = 1-19=89

If Y takes the values 0, 1 and 2, then

PY=0=89×89=6481PY=1=19×89+89×19=1681PY=2=19×19=181

Thus, the probability distribution of X is given by
 

Y P(Y)
0 6481
1 1681
2 181

Page No 32.15:

Question 20:

From a lot containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X if the items are chosen without replacement.

Answer:

Let X denote the number of defective items in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective item=20C425C4=484512650=9692530PX=1=P1 defective item=5C1×20C325C4=570012650=114253PX=2=P2 defective items=5C2×20C225C4=190012650=38253PX=3=P3 defective items=5C3×20C125C4=20012650=4253PX=4=P4 defective items=5C425C4=512650=12530

Thus, the probability distribution of X is given by
 

X P(X)
0 9692530
1 114253
2 38253
3 4253
4 12530

Page No 32.15:

Question 21:

Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine the probability distribution of X.

Answer:

Let X denote the number of hearts in a sample of 3 cards drawn from a well-shuffled deck of 52 cards. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno heart=3952×3952×3952=2764PX=1=P1 heart=1352×3952×3952+3952×1352×3952+3952×3952×1352=2764PX=2=P2 hearts=1352×1352×3952+3952×1352×1352+1352×3952×1352=964PX=3=P3 hearts=1352×1352×1352=164

Thus, the probability distribution of X is given by
 

X P(X)
0 2764
1 2764
2 964
3 164

Page No 32.15:

Question 22:

An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.

Answer:

Let X denote the number of blue balls in a sample of 3 balls drawn from a bag containing 4 red and 3 blue balls. Then, X can take values 0, 1, 2 and 3.
Now,
PX=0=Pno blue ball=47×47×47=64343PX=1=P1 blue ball=37×47×47+47×37×47+47×47×37=144343PX=2=P2 blue balls=37×37×47+47×37×37+37×47×37=108343PX=3=P3 blue balls=37×37×37=27343

Thus, the probability distribution of X is given by
 

X P(X)
0 64343
1 144343
2 108343
3 27343

Page No 32.15:

Question 23:

Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

Answer:

Let X denote the number of spades in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

PX=0=Pno spade=39C252C2=7411326=1934PX=1=P1 spade=13C1×39C152C2=5071326=1334PX=2=P2 spades=13C252C2=781326=117

Thus, the probability distribution of X is given by
 

X P(X)
0 1934
1 1334
2 117

Page No 32.15:

Question 24:

A fair die is tossed twice. If the number appearing on the top is less than 3, it is a success. Find the probability distribution of number of successes.

Answer:

Let X denote the event of getting a number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
PX=0=1636=49PX=1=1636=49 PX=2=436=19

Thus, the probability distribution of X is given by
 

X P(X)
0 49
1 49
2 19

Page No 32.15:

Question 25:

An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represent the number of black balls. What are the possible values of X. Is X a random variable?

Answer:

The possible values of X are 0, 1 and 2, i.e. no black ball, 1 black ball and 2 black balls.
Yes, X is a random variable.
A random variable is a real valued function having domain as the sample space associated with a random experiment.

Page No 32.15:

Question 26:

Let X represent the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are possible values of X?

Answer:

Given: X = Number of heads − Number of tails
 

Number of heads Number of heads Number of heads − Number of tails
0 6 −6
1 5 −4
2 4 −2
3 3 0
4 2 2
5 1 4
6 0 6


Therefore, the possible values of X are :
−6, −4, −2, 0, 2, 4, 6

Page No 32.15:

Question 27:

From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

Answer:

Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non-defectives.Then X can take values 0, 1, 2.
Now,
PX=0=Pno defective bulb=7C210C2=715PX=1=P1 defective bulb=3C1×7C110C2=715PX=2=P2 defective bulbs=3C210C2=115

Thus, the probability distribution of X is given below,
 

X P(X)
0 715
1 715
2 115

Page No 32.15:

Question 28:

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.                                                                                                                  [NCERT EXEMPLAR]

Answer:

As, four balls are to be drawn without replacement and X denotes the number of red balls drawn

So, X is a random variable that can take values 0, 1, 2, 3 or 4

Now,

PX=0=PAll white balls=PWWWW=412×311×210×19=1495,PX=1=POne red balls and three white balls=PRWWW+PWRWW+PWWRW+PWWWR=812×411×310×29+412×811×310×29+412×311×810×29+412×311×210×89=4×8495=32495,PX=2=PTwo red balls and two white balls=PRRWW+PRWRW+PRWWR+PWWRR+PWRWR+PWRRW=812×711×410×39+812×411×710×39+812×411×310×79+412×311×810×79+412×811×310×79+412×811×710×39=6×28495=168495,PX=3=PThree red balls and one white ball=PRRRW+PRRWR+PRWRR+PWRRR=812×711×610×49+812×711×410×69+812×411×710×69+412×811×710×69=4×56495=224495,PX=4=PAll red balls=PRRRR=812×711×610×59=70495

So, the probability distribution of X is as follows:
 

X 0 1 2 3 4
P(X) 1495 32495 168495 224495 70495
 

Page No 32.15:

Question 29:

The probability distribution of a random variable X is given below:
 

X 0 1 2 3
P(X) k k2 k4 k8

(i) Determine the value of k

(ii) Determine P(X  2) and P(X > 2)

(iii) Find P(X  2) + P(X > 2)

Answer:

We have,

The probability distribution of a random variable X is given below:
 

X 0 1 2 3
P(X) k k2 k4 k8

i As, pi=1k+k2+k4+k8=18k+4k+2k+k8=115k8=1 k=815ii As, PX2=1-PX=3=1-k8=1-815×8=1-115=1415Also, PX>2=PX=3=815×8=115iii PX2+PX>2=1415+115              Using ii=1515=1



Page No 32.16:

Question 30:

Let, X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
PX = x = kx,if x = 0 or 12 kx,if x = 2k5-x,if x = 3 or 40,if x > 4
where k is a positive constant. Find the value of k. Also find the probability that you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges.

Answer:

The probability distribution of X is

X 0 1 2 3 4
P(X) 0 k 4k 2k k

The given distribution is a probability distribution.
pi=1
⇒ 0 + k + 4k + 2k + k = 1
⇒8k = 1
k = 0.125

(i) P(getting admission in exactly one college) = P(X = 1) = k = 0.125

(ii) P(getting admission in at most 2 colleges) = P( X ≤ 2) = 0 + k + 4k = 5k = 0.625

(iii) P(getting admission in atleast 2 colleges) = P( X ≥ 2) = 4k + 2k + k = 7k = 0.875



Page No 32.42:

Question 1:

Find the mean and standard deviation of each of the following probability distributions:
(i)

xi : 2 3 4
pi : 0.2 0.5 0.3
                                                                                                                   [NCERT EXEMPLAR]
(ii)
xi : 1 3 4 5
pi : 0.4 0.1 0.2 0.3

(iii)
xi : −5 −4 1 2
pi : 14 18 12 18

(iv)
xi : −1 0 1 2 3
pi : 0.3 0.1 0.1 0.3 0.2

(v)
xi : 1 2 3 4
pi : 0.4 0.3 0.2 0.1

(vi)
xi : 0 1 3 5
pi : 0.2 0.5 0.2 0.1

(vii)
xi : −2 −1 0 1 2
pi : 0.1 0.2 0.4 0.2 0.1

(viii)
xi : −3 −1 0 1 3
pi : 0.05 0.45 0.20 0.25 0.05

(ix)
xi : 0 1 2 3 4 5
pi : 16 518 29 16 19 118
                                                                                                                                                                        [NCERT EXEMPLAR]

Answer:

 (i)

xi pi pixi pixi2
2 0.2 0.4 0.8
3 0.5 1.5 4.5
4 0.3 1.2 4.8
    pixi = 3.1 pixi2 = 10.1

Mean=pixi=3.1Variance=pixi2-Mean2            =10.1-3.12            =10.1-9.61            =0.49Step Deviation=Variance                     =0.49                     =0.7


(ii)
xi pi pixi pixi2
1 0.4 0.4 0.4
3 0.1 0.3 0.9
4 0.2 0.8 3.2
5 0.3 1.5 7.5
    pixi = 3 pixi2 = 12

Mean=pixi=3Variance=pixi2-Mean2            =12-32            =3Step Deviation=Variance                     =3                     =1.732


(iii)
xi pi pixi pixi2
-5 14 -54 254
-4 18 -48 168
1 12 12 12
2 18 28 48
    pixi = -1 pixi2 = 748

Mean=pixi=-1Variance=pixi2-Mean2            =748--12            =9.25-1            =8.25Step Deviation=Variance                     =8.25                     =2.872


(iv)
xi pi pixi pixi2
-1 0.3 -0.3 0.3
0 0.1 0 0
1 0.1 0.1 0.1
2 0.3 0.6 1.2
3 0.2 0.6 1.8
    pixi = 1 pixi2 = 3.4

Mean=pixi=1Variance=pixi2-Mean2            =3.4-1            =2.4Step Deviation=Variance                     =2.4                     =1.549


(v)
xi pi pixi pixi2
1 0.4 0.4 0.4
2 0.3 0.6 1.2
3 0.2 0.6 1.8
4 0.1 0.4 1.6
    pixi = 2 pixi2 = 5

Mean=pixi=2Variance=pixi2-Mean2            =5-22            =1Step Deviation=Variance                     =1                     =1


(vi)
xi pi pixi pixi2
0 0.2 0 0
1 0.5 0.5 0.5
3 0.2 0.6 1.8
5 0.1 0.5 2.5
    pixi = 1.6 pixi2= 4.8

Mean=pixi=1.6Variance=pixi2-Mean2            =4.8-1.62            =2.24Step Deviation=Variance                     =2.24                     =1.497


(vii)
xi pi pixi pixi2
-2 0.1 -0.2 0.4
-1 0.2 -0.2 0.2
0 0.4 0 0
1 0.2 0.2 0.2
2 0.1 0.2 0.4
    pixi = 0 pixi2 =1.2

Mean=pixi=0Variance=pixi2-Mean2            =1.2-02            =1.2Step Deviation=Variance                                 =1.2                     =1.095


(viii)
xi pi pixi pixi2
-3 0.05 -0.15 0.45
-1 0.45 -0.45 0.45
0 0.20 0 0
1 0.25 0.25 0.25
3 0.05 0.15 0.45
    pixi = -0.2 pixi2 =1.6

Mean=pixi=-0.2Variance=pixi2-Mean2            =1.6--0.22            =1.56Step Deviation=Variance                     =1.56                     =1.249


(ix)
xi pi pixi pixi2
0 16 0 0
1 518 518 518
2 29 49 89
3 16 12 32
4 19 49 169
5 118 518 2518
    pixi = 3518 pixi2 = 356

Mean=pixi=3518Variance=pixi2-Mean2            =356-35182            =356-1225324            =1890-1225324            =665324Step Deviation=Variance                     =665324                     =66518



Page No 32.43:

Question 2:

A discrete random variable X has the probability distribution given below:
 

X: 0.5 1 1.5 2
P(X): k k2 2k2 k

(i) Find the value of k.

(ii) Determine the mean of the distribution.                                                                                                           [NCERT EXEMPLAR]

Answer:

The probability distribution of X is given as:
 

X: 0.5 1 1.5 2
P(X): k k2 2k2 k

i As, pi=1k+k2+2k2+k=13k2+2k-1=03k2+3k-k-1=03kk+1-1k+1=03k-1k+1=0k=13 or k=-1but k cannot be negative k=13ii Mean=pixi=0.5×k+1×k2+1.5×2k2+2×k=0.5×13+1×132+1.5×2132+2×13=0.53+19+39+23=1.5+1+3+69=11.59=11590=2318

Page No 32.43:

Question 3:

Find the mean variance and standard deviation of the following probability distribution

xi: a b
pi : p q
where p + q = 1

Answer:

 

xi pi pixi pixi2
a p ap a2p
b q bq b2q
    pixi = ap + bq pixi2= a2p + b2q

Now,Mean=pixi=ap+bqVariance=pixi2-Mean2=a2p+b2q-ap+bq2                                        =a2p+b2q-a2p2-b2q2-2abpq                                        =a2p-a2p2+b2q-b2q2-2abpq                                        =a2p1-p+b2q1-q-2abpq                                        =a2pq+b2qp-2abpq                           p+q=1                                        =pqa2+b2-2ab                                        =pqa-b2Step Deviation=Variance                     =pqa-b2                     =a-bpq

Page No 32.43:

Question 4:

Find the mean and variance of the number of tails in three tosses of a coin.                                                              [NCERT EXEMPLAR]

Answer:

Let X denote the number of tails in three tosses of a coin.

Then, X can take the values 0, 1, 2 and 3.

Now,

PX=0=PHHH=18,PX=1=PTHH or HHT or HTH=38,PX=2=PTTH or THT or HTT=38,PX=3=PTTT=18

Thus, the probability distribution of X is given by
 

X P(X)
0 18
1 38
2 38
3 18

Computation of mean and variance
 
xi pi pixi pixi2
0 18 0 0
1 38 38 38
2 38 68 128
3 18 38 98
    pixi = 32 pixi2 = 3

Mean=pixi=32Variance=pixi2-Mean2           =3-322           =3-94           =34

Page No 32.43:

Question 5:

Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.

Answer:

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno king=48C252C2=11281326=188221PX=1=P1 king=4C1×48C152C2=1921326=32221PX=2=P2 kings=4C252C2=61326=1221

Thus, the probability distribution of X is given by
 

X P(X)
0 188221
1 32221
2 1221


Computation of mean and variance
 
xi pi pixi pixi2
0 188221 0 0
1 32221 32221 32221
2 1221 2221 4221
    pixi = 34221 pixi2 = 36221

Mean=pixi=34221Variance=pixi2-Mean2            =36221-342212          =7956-115648841          =680048841          =4002873

Page No 32.43:

Question 6:

Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.

Answer:

Let X denote the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PHHH=18, PX=1=PTHH or HHT or HTH=38PX=2=PTTH or THT or HTT=38, PX=3=PTTT=18

Thus, the probability distribution of X is given by
 

X P(X)
0 18
1 38
2 38
3 18


Computation of mean and step deviation
 
xi pi pixi pixi2
0 18 0 0
1 38 38 38
2 38 68 128
3 18 38 98
    pixi32 pixi2 = 3

Mean=pixi=32Variance=pixi2-Mean2           =3-322           =34Step Deviation=Variance                     =34                     =0.87

Page No 32.43:

Question 7:

Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.

Answer:

Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno bad egg=10C312C3=120220=611PX=1=P1 bad egg=2C1×10C212C3=90220=922PX=2=P2 bad eggs=2C2×10C112C3=10220=122

Thus, the probability distribution of X is given by
 

X P(X)
0 611
1 922
2 122


Computation of mean
 
xi pi pixi
0 611 0
1 922 922
2 122 111
    pixi = 12

Mean=pixi=12

Page No 32.43:

Question 8:

A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Answer:

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.

Thus, the probability distribution of X is given by
 

X P(X)
1 1136
2 936
3 736
4 536
5 336
6 136


Computation of mean and variance
 
xi pi pixi pixi2
1 1136 1136 1136
2 936 1836 1
3 736 2136 6336
4 536 2036 8036
5 336 1536 7536
6 136 636 1
    pixi = 9136=2.5 pixi2 = 30136=8.4

Mean=pixi=2.5Variance=pixi2-Mean2=8.4-6.25=2.15

Page No 32.43:

Question 9:

A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.

Answer:

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTT, HTTT, THHH, ...

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when a coin is tossed 4 times, we can get minimum 0 and maximum 4 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=416PX=2=P2 heads=616PX=3=P3 heads=416PX=4=P4 heads=116

Thus, the probability distribution of X is given by
 

X P(X)
0 116
1 416
2 616
3 416
4 116


Computation of mean and variance
 
xi pi pixi pixi2
0 116 0 0
1 416 416 416
2 616 1216 2416
3 416 1216 3616
4 116 416 1
    pixi = 2 pixi2 = 5

Mean=pixi=2Variance=pixi2-Mean2            =5-4            =1

Page No 32.43:

Question 10:

A fair die is tossed. Let X denote twice the number appearing. Find the probability distribution, mean and variance of X.

Answer:

Let X denote the event of getting twice the number. Then, X can take the values 2, 4, 6, 8, 10 and 12.

Thus, the probability distribution of X is given by
 

X P(X)
2 16
4 16
6 16
8 16
10 16
12 16


Computation of mean and variance
 
xi pi pixi pixi2
2 16 26 46
4 16 46 166
6 16 66 366
8 16 86 646
10 16 106 1006
12 16 126 1446
    pixi = 7 pixi2 = 3646

Mean=pixi=7Variance=pixi2-Mean2            =60.7-49            =11.7

Page No 32.43:

Question 11:

A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.

Answer:

Let X  be 1 for the appearance of odd numbers 1, 3 or 5 on the die. Then,
PX=1=36=12

Let X be 3 for the appearance of even numbers 2, 4 or 6 on the die. Then,
PX=3=36=12

Thus, the probability distribution of X is given by
 

X P(X)
1 12
3 12


Computation of mean and variance
 
xi pi pixi pixi2
1 12 12 12
3 12 32 92
    pixi = 2 pixi2= 5

Mean=pixi=2Variance=pixi2-Mean2           =5-4           =1

Page No 32.43:

Question 12:

A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

Answer:

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH and TTTT .

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when the coin is tossed 4 times, we can get maximum 4 and minimum 0 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=716PX=2=P2 heads=516PX=3=P3 heads=216PX=4=P4 heads=116

Thus, the probability distribution of X is given by
 

X P(X)
0 116
1 716
2 516
3 216
4 116


Computation of mean and variance
 
xi pi pixi pixi2
0 116 0 0
1 716 716 716
2 516 1016 2016
3 216 616 1816
4 116 416 1
    pixi = 2716 pixi2 = 6116

Mean=pixi=2716=1.7Variance=pixi2-Mean2            =6116-729256            =247256            =0.9

Page No 32.43:

Question 13:

Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

Answer:

There are 5 cards numbered 1, 1, 2, 2 and 3.

Let:
X = Sum of two numbers on cards = 2, 3, 4, 5

Y = Maximum of two numbers = 1, 2, 3


Thus, the probability distribution of X is given by
 

X P(X)
2 110
3 410
4 310
5 210
 

Total number of cards  = 2 cards with "1" on them + 2 cards with "2" on them + 1 card with "3" on it
                                      = 5

Total number of possible choices (in drawing the two cards)= Number of ways in which the two cards can be drawn from the total 5
                                                                                                 =5C2
                                                                                                 =5!2!3!
                                                                                                 = 10

                                                                                                

Probabilty that the two cards drawn would be having the numbers 1 and 1 on them is P(11).

P(11) = C22×C02 ×C01C25
         =1×1×110=110

 

Probabilty that the two cards drawn would be having the numbers 1 and 2 on them is P(11).

P(11) = C22×C02 ×C01C25
         =1×1×110=110

 

  •  
"1" and "2" on them

 

 

⇒ P(12) =
2C1 × 2C1 × 1C0
5C2
   
  "1's" "2's" "3's" Total
Available 2 2 1 5
To Choose 1 1 0 2
Choices 2C1 2C1 1C0 5C2
 

=
2
1
×
2
1
× 1
10
  =
2 × 2 × 1
10
  =
4
10
  =
2
5

 

 

  •  
"1" and "3" on them

 

 

⇒ P(13) =
2C1 × 2C0 × 1C1
5C2
   
  "1's" "2's" "3's" Total
Available 2 2 1 5
To Choose 1 0 1 2
Choices 2C1 2C0 1C1 5C2
 

=
2
1
× 1 ×
1
1
10
  =
2 × 1 × 1
10
  =
2
10
  =
1
5

 

 

  •  
"2" and "2" on them

 

 

⇒ P(22) =
2C0 × 2C2 × 1C1
5C2
   
  "1's" "2's" "3's" Total
Available 2 2 1 5
To Choose 0 2 0 2
Choices 2C0 2C2 1C0 5C2
 

=
1 × 1 × 1
10
  =
1
10

 

 

  •  
"2" and "3" on them

 

 

⇒ P(23) =
2C0 × 2C1 × 1C1
5C2
   
  "1's" "2's" "3's" Total
Available 2 2 1 5
To Choose 0 1 1 2
Choices 2C0 2C1 1C1 5C2
 

=
1 ×
2
1
× 1
10
  =
1 × 2 × 1
10
  =
2
10
  =
1
5

 

Probability for the sum of the numbers on the cards drawn to be

 

  •  
2 ⇒ P(X = 2) = P(11)
    =
1
10
  •  
3 ⇒ P(X = 3) = P(12)
    =
4
10
  •  
4 ⇒ P(X = 4) = P(13 or 22) i.e. P(13 ∪ 22)
    = P(13) + P(22)
    =
2
10
+
1
10
    =
3
10
  •  
5 ⇒ P(X = 2) = P(23)
    =
2
10

 

The probabilty distribution of "x" would be

x 2 3 4 5
P(X = x)
1
10
4
10
3
10
2
10

 

Calculations for Mean and Standard Deviations

 

  x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
  2
1
10
2
10
4
4
10
  3
4
10
12
10
9
36
10
  4
3
10
12
10
16
48
10
  5
2
10
10
10
25
50
10
Total   1
36
10
 
138
10
      = 3.6   = 13.8

 

The Expected Value of the sum


⇒ Expectation of "x"

⇒ E (x) = Σ px
  = 3.6
Variance of the sum of the numbers on the cards
⇒ var (x) = E (x2) − (E(x))2
⇒ var (x) = Σ px2 − (Σ px)2
  = 13.8 − (3.6)2
  = 13.8 − 12.96
  = 0.84
Standard Deviation of the sum of the numbers on the cards
⇒ SD (x) = + Var (x)
  = + 0.84
  = + 0.917

 


Computation of mean and variance
 
xi pi pixi pixi2
2 110 210 410
3 410 1210 3610
4 310 1210 4810
5 210 1 5010
    pixi = 3610=3.6 pixi2 = 13810=13.8

Mean=pixi=3.6Variance=pixi2-Mean2=13.8-12.96=0.84


Thus, the probability distribution of Y is given by
 
Y P(Y)
1 0.1
2 0.5
3 0.4


Computation of mean and variance
 
yi pi piyi piyi2
1 0.1 0.1 0.1
2 0.5 1 2
3 0.4 1.2 3.6
    pixi = 2.3 pixi2 = 5.7

Mean=piyi=2.3Variance=piyi2-Mean2=5.7-5.29=0.41

Page No 32.43:

Question 14:

A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.

Answer:

It is given that "success" denotes the event of getting the numbers 1, 3 or 5. Then,
Psuccess=12

Also, "failure" denotes the event of getting the numbers 2, 4 or 6. Then,
Pfailure=12

Let X denote the event of getting success.Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno success=12×12=14PX=1=P1 success=12×12+12×12=12PX=2=P2 success=12×12=14

Thus, the probability distribution of X is given by
 

X P(X)
0 14
1 12
2 14


Computation of mean and variance
 
xi pi pixi pixi2
0 14 0 0
1 12 12 12
2 14 12 1
    pixi = 1 pixi2 = 32

Mean=pixi=1Variance=pixi2-Mean2=32-1=12

Page No 32.43:

Question 15:

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Answer:

Let X denote the number of defective bulbs in a sample of 3 bulbs drawn from a bag containing 13 bulbs. 5 bulbs in the bag turn out to be defective. Then, X can take the values 0, 1, 2 and 3.

Now,
PX=0=Pno defective bulb=8C313C3=56286=28143PX=1=P1 defective bulb=5C1×8C213C3=140286=70143PX=2=P2 defective bulbs=5C2×8C113C3=80286=40143PX=3=P3 defective bulbs=5C313C3=10286=5143

Thus, the probability distribution of X is given by
 

X P(X)
0 28143
1 70143
2 40143
3 5143

Page No 32.43:

Question 16:

In roulette, Figure, the wheel has 13 numbers 0, 1, 2, ...., 12 marked on equally spaced slots. A player sets Rs 10 on a given number. He receives Rs 100 from the organiser of the game if the ball comes to rest in this slot; otherwise he gets nothing. If X denotes the player's net gain/loss, find E (X).
Figure

Answer:

The wheel has 13 numbers, i.e. 0, 1, 2, ... ,12 marked on equally spaced slots.

∴ Probability of ball resting on any particular number = 113

Let the player set Rs 10 on a given number k.

P(player receives Rs 100) = P(ball rests on it) = 113

X denotes the player's net gain or loss. If he gets the required number, then his gain is Rs 90 (100-10).

If the ball does not rest on the number, then it rests on any of the other 12 numbers. In that case, the player's loss is Rs 10.


Thus, the probability distribution of X is given by

Computation of mean
 

xi pi pixi
90 113 9013
-10 1213 -12013
    pixi = -3013

Mean=Ex=pixi=-3013



Page No 32.44:

Question 17:

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution.                                                                                                                          [CBSE 2014]

Answer:

Let X denotes the number of red cards drawn.

Then, X can take the values 0, 1, 2 or 3.

Now,

PX=0=PBBB=2652×2551×2450=217,PX=1=PRBB or BRB or BBR=3×2652×2651×2550=1334,PX=2=PRRB or RBR or BRR=3×2652×2551×2650=1334,PX=3=PRRR=2652×2551×2450=217

Thus, the probability distribution of X is given by
 

X P(X)
0 217
1 1334
2 1334
3 217

Mean=pixi=0×217+1×1334+2×1334+3×217=0+1334+2634+617=5134=32=1.5

Disclaimer: The answer given in the book is incorrect. The same has been corrected here.

Page No 32.44:

Question 18:

An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.                      [CBSE 2015]

Answer:

As, X represent the number of black balls drawn.

So, it can take values 0, 1 and 2. Yes, X is a random variable.

Now,

PX=0=PRR=57×46=1021,PX=1=PRB or BR=2×57×26=1021,PX=2=PBB=27×16=121Mean=pixi=0×1021+1×1021+2×121=1021+221=1221=47Also, pixi2=02×1021+12×1021+22×121=1021+421=1421=23So, variance=pixi2-Mean2=23-472=23-1649=98-48147=50147

Page No 32.44:

Question 19:

Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.                                                                       [CBSE 2015]

Answer:

As, X denote the larger of the two numbers obtained.

So, X can take the values 3, 4, 5, 6 and 7.

Now,

PX=3=P2,3 or 3,2=2×16×15=115,PX=4=P2,4 or 4,2 or 3,4 or 4,3=4×16×15=215,PX=5=P2,5 or 5,2 or 3,5 or 5,3 or 4,5 or 5,4=6×16×15=15,PX=6=P2,6 or 6,2 or 3,6 or 6,3 or 4,6 or 6,4 or 5,6 or 6,5=8×16×15=415,PX=7=P2,7 or 7,2 or 3,7 or 7,3 or 4,7 or 7,4 or 5,7 or 7,5 or 6,7 or 7,6=10×16×15=13

The probability distribution of X is as follows:
 

X: 3 4 5 6 7
P(X): 115 215 15 415 13

So, Mean, EX=3×115+4×215+5×15+6×415+7×13=315+815+1515+2415+3515=8515=173Also, EX2=32×115+42×215+52×15+62×415+72×13=915+3215+7515+14415+24515=50515=1013VarX=EX2-EX2=1013-1732=1013-2899=303-2899=149

Page No 32.44:

Question 20:

In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses.

Answer:


The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equatl to 4 in the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P (X = 5) = P(Getting number greater than 4 in first throw) = 13
​P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) = 46×26=29
​P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) = 46×46×26=427
​P (X = -3) = P(Getting number less than equal to 4 in all three throws) = 46×46×46=827
 

5 4 3 -3
P(X) 13 29 427 827

E (X) = 5×13+4 29+3427-3 827=12745+24+12-24=5727
Expected value of the amount he wins/loses is 5727.



Page No 32.45:

Question 1:

Write the values of 'a' for which the following distribution of probabilities becomes a probability distribution:

X = xi : −2 −1 0 1
P (X = xi) : 1-a4 1+2a4 1-2a4 1+a4

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) = 1

1-a4+1+2a4+1-2a4+1+a4=144=11=1Now, 01-a41 01-a4 -1-a3 -3a1       ...101+a41 01+a4 -1a3       ...201-2a41 01-2a4 -1-2a3 -32a12       ...301+2a41 01+2a4 -12a3 -12a32       ...4From 1, 2, 3 and 4, we get-12a12

Page No 32.45:

Question 2:

For what value of k the following distribution is a probability distribution?

X = xi : 0 1 2 3
P (X = xi) : 2k4 3k2 − 5k3 2k − 3k2 3k − 1

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

2k4+3k2-5k3+2k-3k2+3k-1=12k4-5k3+5k=22k4-5k3+5k-2=0k-1k-22k2+k-1=0k-1k-22k-1k+1=0k=-1 , 12, 1, 2Neglecting -1 , 1 and 2 as they give the value of probability negative or greater than 1

∴ k =12

Page No 32.45:

Question 3:

If X denotes the number on the upper face of a cubical die when it is thrown, find the mean of X.

Answer:

A cubical die can show 1, 2, 3, 4, 5 or 6 on its face.
 

xi pi pixi
1 16 16
2 16 26
3 16 36
4 16 46
5 16 56
6 16 66


Mean = pixi = 16+26+36+46+56+66=216=3.5



Page No 32.46:

Question 4:

If the probability distribution of a random variable X is given by

X = xi : 1 2 3 4
P (X = xi) : 2k 4k 3k k
Write the value of k.

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

2k+4k+3k+k=110k=1k=110=0.1

Page No 32.46:

Question 5:

Find the mean of the following probability distribution:

X = xi : 1 2 3
P (X = xi) : 14 18 58

Answer:

xi pi pixi
1 14 14
2 18 28
3 58 158


Mean = pixi = 14+28+158=2+2+158=198

Page No 32.46:

Question 6:

If the probability distribution of a random variable X is as given below:

X = xi : 1 2 3 4
P (X = xi) : c 2c 4c 4c
Write the value of P (X ≤ 2).

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

c+2c+4c+4c=111c=1c=111Now,PX2=PX=1+PX=2=110+210=311

Page No 32.46:

Question 7:

A random variable has the following probability distribution:

X = xi : 1 2 3 4
P (X = xi) : k 2k 3k 4k
Write the value of P (X ≥ 3).

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

k+2k+3k+4k=110k=1k=110Now, PX3=PX=3+PX=4=310+410=710

Page No 32.46:

Question 1:

If a random variable X has the following probability distribution:

X : 0 1 2 3 4 5 6 7 8
P (X) : a 3a 5a 7a 9a 11a 13a 15a 17a
then the value of a is

(a) 781

(b) 581

(c) 281

(d) 181

Answer:

(d) 181


We know that the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

a+3a+5a+7a+9a+11a+13a+15a+17a=181a=1a=181

Page No 32.46:

Question 2:

A random variable X has the following probability distribution:

X : 1 2 3 4 5 6 7 8
P (X) : 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events E = {X : X is a prime number}, F = {X : X < 4}, the probability P (EF) is

(a) 0.50
(b) 0.77
(c) 0.35
(d) 0.87

Answer:

(b) 0.77
 
E = {X is a prime number}  = {2, 3, 5, 7}

PE =P2+P3+P5+P7= 0.62F= {X<4}  = {X=1,2,3} PF= P1+P2+P3=0.5Now,EF = {2,3}PEF=P2+P3=0.35PEF=PE+PF-PEF=0.62+0.50-0.35=0.77

Page No 32.46:

Question 3:

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P (X = 3) = 2 P (X = 1) and P (X = 2) = 0.3, then P (X = 0) is
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4

Answer:

(d) 0.4

Let:
P(X = 0) = m
P(X = 1) = k.

Now,
P(X = 3) = 2k
 

xi pi pixi
0 m 0
1 k k
2 0.3 0.6
3 2k 6k


Mean = pixi

0+k+0.6+6k=1.37k=1.3-0.6k=0.77=0.1

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)  = 1

m+0.1+0.3+0.2=1m+0.6=1m=0.4


 



Page No 32.47:

Question 4:

A random variable has the following probability distribution:

X = xi : 0 1 2 3 4 5 6 7
P (X = xi) : 0 2 p 2 p 3 p p2 2 p2 7 p2 2 p
The value of p is

(a) 1/10
(b) −1
(c) −1/10
(d) 1/5

Answer:

(a) 1/10

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

0+2p+2p+3p+p2+2p2+7p2+2p=110p2+9p-1=010p-1p+1=0p=110 or -1            Neglecting -1 as the value of the probabilitiy cannot be negative

Page No 32.47:

Question 5:

If X is a random-variable with probability distribution as given below:

X = xi : 0 1 2 3
P (X = xi) : k 3 k 3 k k
The value of k and its variance are

(a) 1/8, 22/27
(b) 1/8, 23/27
(c) 1/8, 24/27
(d) 1/8, 3/4

Answer:

(d) 1/8, 3/4

We know that the sum of probabilities in a probability distribution is always 1.

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

k+3k+3k+k=18k=1k=18

Now,
 

xi pi pixi pixi2
0 k = 18 0 0
1 3k = 38 38 38
2 3k = 38 68 128
3 k = 18 38 98
    pixi = 128=32 pixi2 = 248

Mean=pixi=32Variance=pixi2-Mean2=248-322=248-94=24-188=68=34

Page No 32.47:

Question 6:

Mark the correct alternative in the following question:

The probability distribution of a discrete random variable X is given below:
 

X: 2 3 4 5
P(X): 5k 7k 9k 11k

The value of k is

(a) 8                                (b) 16                                (c) 32                                (d) 48

Answer:

The probability distribution of a discrete random variable X is given below:
 

X: 2 3 4 5
P(X): 5k 7k 9k 11k

As, pi=15k+7k+9k+11k=132k=1 k=32

Disclaimer: The question is incorrect, it should be "k" instead "E(X)". The same has been corrected here.

Page No 32.47:

Question 7:

Mark the correct alternative in the following question:

For the following probability distribution:
 

X: 4 3 2 1 0
P(X): 0.1 0.2 0.3 0.2 0.2

The value of E(X) is

(a) 0                                (b) 1                                (c) 2                                (d) 1.8

Answer:

The probability distribution of X is given below:
 

X: 4 3 2 1 0
P(X): 0.1 0.2 0.3 0.2 0.2

EX=-4×0.1+-3×0.2+-2×0.3+-1×0.2+0×0.2=-0.4-0.6-0.6-0.2=-1.8

Hence, the correct alternative is option (d).

Page No 32.47:

Question 8:

Mark the correct alternative in the following question:

For the following probability distribution:
 

X: 1 2 3 4
P(X): 110 15 310 25

The value of E(X2) is

(a) 3                                (b) 5                                (c) 7                                (d) 10

Answer:

The probability distribution of X is given below:
 

X: 1 2 3 4
P(X): 110 15 310 25

EX2=12×110+22×15+32×310+42×25=110+810+2710+6410=10010=10

Hence, the correct alternative is option (d).

Page No 32.47:

Question 9:

Mark the correct alternative in the following question:

Let X be a discrete random variable. Then the variance of X is

(a) E(X2)                            (b) E(X2) + (E(X))2                            (c) E(X2- (E(X))2                            (d) EX2-EX2

Answer:

Since, the variance of a discrete random variable X is given by:

Var(X) = E(X2- (E(X))2


Hence, the correct alternative is option (c).



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