Question

A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads $90\%$ of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probability that it is the unbiased coin, is:

• A. $\frac{5}{12}$
• B. $\frac{5}{24}$
• C. $\frac{1}{3}$
• D. $\frac{3}{8}$

Answer 1

Answer: (B)

$\frac{5}{24}$

IF $E_1,E_2,E_3$ be $n$ mutually exclusive and exhaustive events and $A$ is an event which occurs together with either $E_1,E_2,E_3,....E_n$ from a portion of the sample space $S$ and $A$ be any event, then

$P\left(\frac{E_k}{A}\right)=\frac{P\left(E_k\right).P\left(\frac{A}{E_k}\right)}{P\left(E_1\right).P\left(\frac{A}{E_1}\right).P\left(E_2\right).\left(\frac{A}{E_2}\right).+....+P\left(E_k\right).\left(\frac{A}{E_k}\right)}$,

where $S=A_1\cup A_2\cup ....A_k$ and $S$ is a sample space

$P\left(\frac{\text{Heads}}{C}\right)=\frac{(\frac{1}{3}\times \frac{1}{2})}{(\frac{1}{3}\times 1)+(\frac{1}{3}\times \frac{9}{10})+(\frac{1}{3}\times \frac{1}{2})}=\frac{5}{10+9+5}=\frac{5}{24}$