Question

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white ?

Answer 1

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such that

$P\left(A\right)=P\left(B\right)=P\left(C\right)=P\left(D\right)$

There are 16 balls out of which 11 are not white. Therefore,

$P\left(A\right)=\frac{11}{16}$

Now, required probability = $P(A\cap B\cap C\cap D)$

$=P\left(A\right)P\left(B\right)P\left(C\right)P\left(D\right)=(\frac{11}{16}{)}^4$