Probability 3 Question 38

38. An unbiased coin is tossed. If the result in a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well-shuffled pack of eleven cards numbered 2,3,4,,12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8 ?

(1994,5 M)

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Solution:

  1. Let, E1= the event noted number is 7

E2= the event noted number is 8

H= getting head on coin

T= be getting tail on coin

By law of total probability,

P(E1)=P(H)P(E1/H)+P(T)P(E1/T) and P(E2)=P(H)P(E2/H)+P(T)P(E2/T) where, P(H)=1/2=P(T)

P(E1/H)= probability of getting a sum of 7 on two dice

Here, favourable cases are

(1,6),(6,1),(2,5),(5,2),(3,4),(4,3).

P(E1/H)=636=16

Also, P(E1/T)= probability of getting 7 numbered card out of 11 cards

=111

P(E2/H)= probability of getting a sum of 8 on two dice

Here, favourable cases are

(2,6),(6,2),(4,4),(5,3),(3,5).

P(E2/H)=536

P(E2/T)= probability of getting ’ 8 ’ numbered card out of 11 cards

=1/11

P(E1)=12×16+12×111=112+122=17132

and

P(E2)=12×536+12×111=1291396=91729

Now, E1 and E2 are mutually exclusive events.

Therefore,

P(E1 or E2)=P(E1)+P(E2)=17132+91792=193792



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