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# A bag A contains

Answer :

Let E1 be the event of selecting bag A or getting 1or 2 on die. Let E2 be the event of selecting bag B or getting 3,4,5 or 6 on die. Let A be the event drawing one red and one black ball.

Now,P(A/ E1) = Probability of drawing a red ball and a black ball

when bag A has been chosen.

P(A/ E1) = 4C1. 6C1/ 10C2 = P(A/ E2) = Probability of drawing a red ball and a black ball

when bag B has been chosen.

P(A/ E2) = 7C1. 3C1/ 10C2= Using the law of total probability ,we have

P(A) = P(E1) P(A/ E1)+P(E2) P(A/ E2) OR

Let p = probability of getting head and q = probability of getting tail Let X be the number of heads in the 4 tosses of coin.

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 Now, mean E(X) =  Variance, Var(X) =  Rate this question :

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