Question

Die $A$ has $4$ red and $2$ white faces where as die $B$ has two red and $4$ white faces. $A$ fair coin is tossed. If head turns up, the game continues by throwing die $A$, if tail turns up then die $B$ is to be used. If the first two throws resulted in red, what is the probability of getting red face at the third throw ?

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

Answer 1

Answer: (C)

$\frac{3}{5}$

Let $E_1$ be the event that die $A$ is used and $E_2$ be the event that die $B$ is used

Let $C$ be the event that a red face appears in any throw.
$P\left(E_1\right)=\frac{1}{2}=P\left(E_2\right)$

$P\left(C/E_1\right)=\frac{{}^4{C}_1}{{}^6{C}_1}=\frac{2}{3}$,

$P\left(C/E_2\right)=\frac{{}^2{C}_1}{{}^6{C}_1}=\frac{1}{3}$

$P\left(C\right)=P\left(E_1\right)P\left(C/E_1\right)+P\left(E_2\right).P\left(C/E_2\right)=1/2\times 2/3+1/2\times 1/3=1/2$

Let $D$ be the event that red face appears in third throw
Let $E$ be the even that red face appears in first two throws
$P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

$P\left(E/E1\right)=\frac{2}{3}.\frac{2}{3}={\left(\frac{2}{3}\right)}^2$,

$P\left(D/E{E}_1\right)=\frac{2}{3}$

$P\left(E/E2\right)=\frac{1}{3}\times \frac{1}{3}={\left(\frac{1}{3}\right)}^2$

$P\left(D/E{E}_2\right)=\frac{1}{3}$

$P\left(D/E\right)=\frac{P\left(E_1\right)P\left(E/E_1\right)P\left(D/E{E}_1\right)+P\left(E_2\right).P\left(E/E_2\right)P\left(D/E{E}_2\right)}{P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)}$

$=\frac{1/2(2/3{)}^2\times 2/3+1/2(1/3{)}^2\times 1/3}{1/2\times (2/3{)}^2+1/2\times (1/3{)}^2}=\frac{3}{5}$