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 probablity that :
(i) None is white
(ii) All are white
(iii) Only two are white

Answer 1

$5white,7red,8blackballsi)Probabilitythatnoneiswhite=\frac{15}{20}\times \frac{15}{20}\times \frac{15}{20}\times \frac{15}{20}=({\frac{3}{4})}^{4}ii)Probabilityallarewhite=\frac{5}{20}\times \frac{5}{20}\times \frac{5}{20}\times \frac{5}{20}=({\frac{1}{4})}^{4}iii)Probabilitythatonlytwoiswhite={4_C}_2\times \frac{5}{20}\times \frac{5}{20}\times \frac{15}{20}\times \frac{15}{20}=54({\frac{1}{4})}^4$