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# A card from a pac

Answer :

Let E1 be the event that lost card is a spade. Let E2 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/ E1) = Probability of drawing three spade cards given that

lost card is spade.

P(A/ E1) = 12C3 / 51C3 = P(A/ E2) = Probability of drawing three spade cards given that

lost card is non spade.

P(A/ E2) = 13C3 / 51C3 = 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) = nCr.pr.qn-r when n=4 , r=0,1,2,3,4

P(X=0) = 4C0.p0.q4-0 = 4C0 P(X=1) = 4C1.p1.q4-1 = 4C1 P(X=2) = 4C2.p2.q4-2 = 4C2 P(X=3) = 4C3.p3.q4-3 = 4C3 P(X=4) = 4C4.p4.q4-4 = 4C4 Thus, the probability distribution of X is given by

X: 0 1 2 3 4

P(x): Now, mean E(X) = = =   Rate this question :

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