Question

A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
(i) none is white?
(ii) all are white?
(iii) any two are white?

Answer 1

Let X be the number of white balls drawn when 4 balls are drawn with replacement.
X follows binomial distribution with n = 4.

$$\begin{gathered} p\mathit{}=\mathrm{Probability}\mathrm{for}\mathrm{a}\mathrm{white}\mathrm{ball}=\frac{\mathrm{No}\mathrm{of}\mathrm{white}\mathrm{balls}}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{balls}} \\ =\frac{5}{20} \\ =\frac{1}{4} \\ \text{and}q\mathit{}=1-p=\frac{3}{4} \\ P(X=r)={}^{4}C_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{4-r} \\ \left(\mathrm{i}\right)\mathrm{Prob}\mathrm{that}\mathrm{none}\mathrm{is}\mathrm{white}=P(X=0)={}^{4}C_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{4-0}=\frac{81}{256} \\ \left(\mathrm{ii}\right)\mathrm{Prob}\mathrm{that}\mathrm{all}\mathrm{are}\mathrm{white}=P(X=4)={}^{4}C_4{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^{4-4}=\frac{1}{256} \\ \left(\mathrm{iii}\right)\mathrm{Prob}\mathrm{that}\mathrm{any}\mathrm{two}\mathrm{are}\mathrm{white}=\mathit{}P(X=2)={}^4\mathrm{C}_2{\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^{4-2}=\frac{54}{256}=\frac{27}{128} \end{gathered}$$