Q1 A number of 4 different digits is formed by using 1, 2, 3, 4, 5, 6, 7 - Maths - Probability

Question

Q1 A number of 4 different digits is formed by using 1, 2, 3, 4,
5, 6, 7 Find
the probability that it is divisible by 5

Q2 A bag contains 5 red, 4 blue and an unknown
number of m green balls Two balls are drawn If probability of both
being green is 1/7 find m

Manmeet Singh · Accepted Answer

(1)

The total four digit numbers that can be formed by the digits 1, 2, 3, 4, 5, 6, 7 are 7 × 6 × 5 × 4 = 840

Total four digit numbers that can be formed by the digits which is divisible by 5 are:

For the number to be divisible by 5, the last digit must be 5.

So, the number of such numbers are 6 × 5 × 4 = 120. (it is given in the question that the number should be formed by different digits)

The probability that it is divisible by 5 = $\frac{n u m b e r o f f a v o u r a b l e o u t c o m e s}{t o t a l n u m b e r o f o u t c o m e s} = \frac{120}{840} = \frac{1}{7}$

(2)

The number of red balls = 5, the number of blue balls = 4, the number of green balls = m

Total number of balls = 5 + 4 + m = 9 + m

Since two balls are drawn, the number of ways in which both the balls being drawn are green = ${{}^{m}C}_{2}$

Total number of ways in which two balls can be drawn are ${{}^{9 + m}C}_{2}$

Probability =

$\frac{{{}^{m}C}_{2}}{{{}^{9 + m}C}_{2}} = \frac{\frac{m!}{2! (m - 2)!}}{\frac{(9 + m)!}{2! (9 + m - 2)!}} = \frac{\frac{m!}{2! (m - 2)!}}{\frac{(9 + m)!}{2! (7 + m)!}} = \frac{\frac{m (m - 1) (m - 2)!}{(m - 2)!}}{\frac{(9 + m) (8 + m) (7 + m)!}{(7 + m)!}} = \frac{m (m - 1)}{(9 + m) (8 + m)} \Rightarrow \frac{m (m - 1)}{(9 + m) (8 + m)} = \frac{1}{7} \Rightarrow 7 m (m - 1) = (9 + m) (8 + m) \Rightarrow 7 m^{2} - 7 m = m^{2} + 17 m + 72 \Rightarrow 6 m^{2} - 24 m - 72 = 0 \Rightarrow m^{2} - 4 m - 12 = 0 \Rightarrow m^{2} - 6 m + 2 m - 12 = 0 \Rightarrow (m - 6) (m + 2) = 0 \Rightarrow m = 6, - 2$

since the number of balls can never be negative,

So, m = – 2 is rejected.

Hence, m = 6

Thus, the number of green balls = 6.

1

Q1. A number of 4 different digits is formed by using 1, 2, 3, 4, 5, 6, 7. Find the probability that it is divisible by 5. Q2.A bag contains 5 red, 4 blue and an unknown number of m green balls.Two balls are drawn. If probability of both being green is 1/7 find m.

Q1. A number of 4 different digits is formed by using 1, 2, 3, 4, 5, 6, 7. Find

the probability that it is divisible by 5.

Q2.A bag contains 5 red, 4 blue and an unknown number of m green balls.Two balls are drawn. If probability of both being green is 1/7 find m.