Q) A box contains 90 cards which are numbered from 1 to 90.If one card is drawn at random from the box.Find the probablity that it is .? 

1) A two digit number.

2) A perefect square number.

3) A number divisible by 3.

4) A number divisible by 5.

5) A number divisible by 3or5

6) a number dividible by 3and5

Dear Student,

Please find below the solution to the asked query:

We know formula for probability :

Probability P ( E )  = $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{desired} \mathrm{events} \mathrm{n} ( \mathrm{E} )}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{events} \mathrm{n} ( \mathrm{S} )}$

Here , n ( S ) =  90  ( As box contains 90 cards )

1 )  Total number of cards with two digits number  =  81 (  10 to  90 )

So,

n ( E ) = 81 

Therefore ,

Probability of getting a two digit number  = $\frac{81}{90} = \frac{\mathbf{9}}{\mathbf{10}}$                                      ( Ans )

2 ) Total number of perfect square number  =  9 ( 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 )

So,

n ( E ) = 9

Therefore ,

Probability of getting a perfect square number  = $\frac{9}{90} = \frac{\mathbf{1}}{\mathbf{10}}$                                      ( Ans )

3 ) Total number of numbers divisible by 3  =  30 ( 3 , 6 , 9 , 12 ,15 ,18 , 21, 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48, 51 , 54, 57 , 60 , 63, 66 , 69 , 72 , 75 , 78 , 81 , 84 , 87 , 90 )

So,

n ( E ) = 30

Therefore ,

Probability of getting a  number  divisible by 3 = $\frac{30}{90} = \frac{\mathbf{1}}{\mathbf{3}}$                                      ( Ans )

4 ) Total number of numbers divisible by 5  =  18 ( 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 )
So,

n ( E ) = 18

Therefore ,

Probability of getting a  number  divisible by 3 = $\frac{18}{90} = \frac{\mathbf{1}}{\mathbf{5}}$                                      ( Ans )

5 ) Total number of numbers divisible by 3 or 5  =  42 ( 3 , 5, 6 , 9 , 10 , 12 ,15 ,18 , 20 , 21, 24 , 25 , 27 , 30 , 33 , 35 , 36 , 39 , 40 , 42 , 45 , 48, 50 , 51 , 54 , 55 , 57 , 60 , 63 , 65 , 66 , 69 , 70 , 72 , 75 , 78 , 80 , 81 , 84 , 85 , 87 , 90 )

So,

n ( E ) = 42

Therefore ,

Probability of getting a  number  divisible by 3 or 5  = $\frac{42}{90} = \frac{\mathbf{7}}{\mathbf{15}}$                                      ( Ans )

6 ) Total number of numbers divisible by 3 and  5  =  6 (  15 , 30 , 45  , 60 , 75, 90 )

So,

n ( E ) = 6

Therefore ,

Probability of getting a  number  divisible by 3 and 5  = $\frac{6}{90} = \frac{\mathbf{1}}{\mathbf{15}}$                                      ( Ans )
Hope this information will clear your doubts about probability .

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