CBSE[All India]_XII_Mathematics_2004_Set I

Total no. of tickets in a bag = 20

2 tickets can be drawn out of 20 in ²⁰C₂ ways

So exhaustive no. of cases = ²⁰C₂= (20)(19)/2 = 190 ways

(i) there are 8 prime numbers from 1 to 20 (i.e. 2,3,5,7,11,13,17,19)

Out of these 8 prime numbers, 2 prime numbers can be drawn in ⁸C₂ways.

Favorable no. of cases = ⁸C₂= (8) (7)/2 =28

Thus the probability that both the tickets have prime number on them 
= Favorable number of cases / Total number of cases
= 8C2/ 20C2= 28/190 = 14/95

(ii) There are 8 prime numbers from 1 to 20 (i.e. 2,3,5,7,11,13,17,19)  and 5 numbers which are multiple of 4 from 1 to 20

Out of these 8 prime numbers and 5 multiples of 4, 1prime number and 1 multiple of 4 can be drawn in ⁸C₁⁵C₁ways

Favorable no. of cases =⁸C₁⁵C₁=85 = 40 ways

Thus the probability that one of the tickets has prime number and other has a multiple of 4 
= Favorable number of cases / Total number of cases

= ⁸C₁⁵C₁/²⁰C₂= 40/190 = 4/95

Let S be the sample space associated with the experiment when two dice are tossed once. Then,

n(S) = 6² = 36

Let A = Getting a even number on the first dice

B = Getting a total of 8

AB= Getting an even number on the first dice and a total of 8.

A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6,6)}
n(A) = 18

B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

n(B) = 5

AB= {(2, 6), (4, 4), (6, 2)}

n(AB) = 3

P(A) = n(A)/n(S) = 18/36

P(B) = n(B)/n(S)= 5/36

P(AB) =n(AB) /n(S)= 3/36
Required probability:

P(AB) = P(A) + P(B) - P(AB)

18/36 +5/36 -3/36 = 20/36 = 5/9

As X denote the number of defective bulbs in the sample space, therefore X can take the values of 0,1,2,3.

Total numbers of bulb in the lot = 30= 6 defective bulbs +24 non defective bulbs

Probability of getting a defective bulb =  6/30 =1/5

Probability of getting a non defective bulb =  24/30= 4/5

P(X=0)=P( no defective bulbs)