Answer :

Let E_{1} be the event of selecting bag A or getting 1or 2 on die.

⇒

Let E_{2} 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/ E_{1}) = Probability of drawing a red ball and a black ball

when bag A has been chosen.

⇒P(A/ E_{1}) = ^{4}C_{1}. ^{6}C_{1}/ ^{10}C_{2} =

P(A/ E_{2}) = Probability of drawing a red ball and a black ball

when bag B has been chosen.

⇒P(A/ E_{2}) = ^{7}C_{1}. ^{3}C_{1}/ ^{10}C_{2}=

Using the law of total probability ,we have

P(A) = P(E_{1}) P(A/ E_{1})+P(E_{2}) P(A/ E_{2})

**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) = ^{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

Now, mean E(X) =

Variance, Var(X) =

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