A box has 5 balls . Two balls are drawn at random and are found to be white . What is the probability that the box has 4 white balls ?

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$$\begin{gathered} \mathrm{Box} \mathrm{may} \mathrm{have} 2 \mathrm{whites} \mathrm{balls}, 3 \mathrm{whites} \mathrm{balls}, 4 \mathrm{whites} \mathrm{balls} \mathrm{or} 5 \mathrm{whites} \mathrm{balls} \\ \mathrm{Hence} \mathrm{probability} \mathrm{of} \mathrm{getting} 2 \mathrm{white} \mathrm{balls} \mathrm{for} \mathrm{above} \mathrm{cases} \mathrm{will} \mathrm{be}{ \\ }^2{\mathrm{C}}_2{,}^3{\mathrm{C}}_2{,}^4{\mathrm{C}}_2 \mathrm{or}{ }^5{\mathrm{C}}_2 \mathrm{respectively}. \\ \mathrm{Hence} \mathrm{probability} \mathrm{that} \mathrm{box} \mathrm{has} 4 \mathrm{white} \mathrm{balls} \\ =\frac{{}^4{\mathrm{C}}_2}{{}^2{\mathrm{C}}_2{+}^3{\mathrm{C}}_2{+}^4{\mathrm{C}}_2 {+}^5{\mathrm{C}}_2} \\ =\frac{{\displaystyle \frac{4\times 3}{2\times 1}}}{1+3+\frac{4\times 3}{2\times 1}+\frac{5\times 5}{2\times 1}} \\ =\frac{6}{1+3+6+10} \\ =\frac{6}{20} \\ =\frac{3}{10} \end{gathered}$$

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