Question

A purse contains four coins; two coins having been drawn are found to be 2 Rupee coins; find the chance (i) that all coins are 2 Rupee coins (ii) that if the coins are replaced, another drawing will give a 2 Rupee coin.

• A. $\frac{2}{5}$,$\frac{7}{8}$
• B. $\frac{3}{5}$,$\frac{1}{8}$
• C. $\frac{3}{5}$,$\frac{7}{8}$
• D. $\frac{2}{5}$,$\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{3}{5}$,$\frac{7}{8}$

The purse may contain $2$ or $3$ or $4$ two-rupee coins. All these three hypotheses are equally likely i.e., $P_1=P_2=P_3$

Hence $P_1=\frac{2}{4}\times \frac{1}{3}=\frac{1}{6},P_2=\frac{3}{4}\times \frac{2}{3},P_3=\frac{4}{4}\times \frac{3}{3}=1$

$\therefore$ The chance that all are two-rupee coins

$=\frac{p_3}{\sum \left(p\right)}=\frac{1}{\frac{1}{6}+\frac{1}{2}+1}=\frac{6}{10}=\frac{3}{5}$

Again $\frac{p_1}{1}=\frac{p_2}{3}=\frac{p_3}{6}=\frac{\sum \left(p\right)}{10}$

$\therefore Q_1=\frac{1}{10},Q_2=\frac{3}{10},Q_3=\frac{6}{10}$

$\therefore$ The chance that another drawing will give a two rupee coin

$=(Q_1\times \frac{2}{4})+(Q_2\times \frac{3}{4})+(Q_3\times 1)$

$=\frac{1}{20}+\frac{9}{40}+\frac{6}{10}=\frac{7}{8}.$