Answer :

Let E_{1} be the event that lost card is a spade.

⇒

Let E_{2} be the event that lost card is not a spade.

⇒

Let A be the event drawing three spade cards from remaining cards without replacement.

Now,P(A/ E_{1}) = Probability of drawing three spade cards given that

lost card is spade.

⇒P(A/ E_{1}) = ^{12}C_{3} / ^{51}C_{3} =

P(A/ E_{2}) = Probability of drawing three spade cards given that

lost card is non spade.

⇒P(A/ E_{2}) = ^{13}C_{3} / ^{51}C_{3} =

By Baye’s theorem , we have required probability

=

⇒

**OR**

Let p = probability of getting defective bulb =

and q = probability of getting non defective bulb =

Let X be the number of defective bulbs out of 4 drawn.

X can take values 0,1,2,3 and 4.

Since X is a binomial variate with parameters n=4 , such that

P(X=r) = ^{n}C_{r}.p^{r}.q^{n-r} when n=4 , r=0,1,2,3,4

P(X=0) = ^{4}C_{0}.p^{0}.q^{4-0} = ^{4}C_{0}

P(X=1) = ^{4}C_{1}.p^{1}.q^{4-1} = ^{4}C_{1}

P(X=2) = ^{4}C_{2}.p^{2}.q^{4-2} = ^{4}C_{2}

P(X=3) = ^{4}C_{3}.p^{3}.q^{4-3} = ^{4}C_{3}

P(X=4) = ^{4}C_{4}.p^{4}.q^{4-4} = ^{4}C_{4}

Thus, the probability distribution of X is given by

X: 0 1 2 3 4

P(x):

Now, mean E(X) =

=

=

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