Question

A bag contains $5$ white, $7$ red and $8$ black balls. If four balls are drawn one by one with replacement. What is the probability that at least one is white?

Answer 1

Given $5$-white balls

$7$-red balls

$8$-black balls

P(no while)

$=\frac{{}^{15}{C}_1}{{}^{20}{C}_1}\times \frac{{}^{15}{C}_1}{{}^{20}{C}_1}\times \frac{{}^{15}{C}_1}{{}^{20}{C}_1}\times \frac{{}^{15}{C}_1}{{}^{20}{C}_1}=(\frac{15}{20}{)}^4=(\frac{3}{4}{)}^4=\frac{81}{256}$

P (at least one white ball ) = 1-P(no white ball)

$=1-\frac{81}{250}=\frac{125}{256}$