Q. 295.0( 1 Vote )

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) =


=


=




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