A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one buy one with replacement, what is the probability that i. None is white? ii. All are white? iii. Any two are white?

Explanation

Let `A, B, C, D` denote the events of not getting a white ball in first, second, third and fourth draw respectively.

Since the balls are drawn with replacement, therefore, `A, B, C, D ` are independent events such that
`P(A)=P(B)=P(C)=P(D)`.

Since out of `16` balls, `11` are not white, therefore, `P(A)=frac{11}{16}.`

`therefore `Required probability `=P(A). P (B) . P (C). P (D)`


`=frac{11}{16} times frac{11}{16} times frac{11}{16} times frac{11}{16}=(frac{11}{16})^{4}`