A bag A contains

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) = ⁿC_r.p^r.q^n-r when n=4 , r=0,1,2,3,4

P(X=0) = ⁴C₀.p⁰.q^4-0 = ⁴C₀

P(X=1) = ⁴C₁.p¹.q^4-1 = ⁴C₁

P(X=2) = ⁴C₂.p².q^4-2 = ⁴C₂

P(X=3) = ⁴C₃.p³.q^4-3 = ⁴C₃

P(X=4) = ⁴C₄.p⁴.q^4-4 = ⁴C₄

Thus, the probability distribution of X is given by

Now, mean E(X) =

Variance, Var(X) =