Question

The content of $3$ urns $1,2,3$ are as follows : $1$ white, $2$ black, $3$ red balls; $2$ white, $1$ black, $1$ red balls; $4$ white, $5$ black, $3$ red balls. One urn is selected at random and two balls are drawn and these comes out to be white and red. Then the probability that they come from

• A. $2nd\text{urn is}\frac{55}{118}$
• B. $3rd\text{urn is}\frac{30}{118}$
• C. $1st\text{urn is}\frac{33}{118}$
• D. $3rd\text{urn is}\frac{59}{118}$

Answer 1

Answer: (A), (B), (C)

$1st\text{urn is}\frac{33}{118}$

Let $E_i$ be the event of choosing urn '$i$', ($i=1,2,3$). Let $E$ be the event of getting one white and one red ball when $2$ balls are drawn from the chosen urn.
$P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{3}$and$P\left(E_3\right)=\frac{1}{3}$
$P\left(\frac{E}{E_1}\right)=\frac{1\times 3}{{}^6{C}_2},P\left(\frac{E}{E_2}\right)=\frac{2\times 1}{{}^4{C}_2}$and$P\left(\frac{E}{E_3}\right)=\frac{4\times 3}{{}^{12}{C}_2}$
Using total probability theorem, we have:
$P\left(E\right)=\sum P\left(E_i\right).P\left(E/E_i\right)=\frac{1}{3}[\frac{1\times 3}{{(}^6{C}_2)}+\frac{2\times 1}{{(}^4{C}_2)}+\frac{4\times 3}{{(}^{12}{C}_2)}]=\frac{1}{3}[\frac{1}{5}+\frac{1}{3}+\frac{2}{11}]$
Using Baye's therorem
$\Rightarrow P\left(E_1/E\right)=\frac{\frac{1}{3}\left(\frac{1}{5}\right)}{\frac{1}{3}(\frac{1}{5}+\frac{1}{3}+\frac{2}{11})}=\frac{33}{33+55+30}=\frac{33}{118}$
$\Rightarrow P\left(E_2/E\right)=\frac{\frac{1}{3}\left(\frac{1}{3}\right)}{\frac{1}{3}(\frac{1}{5}+\frac{1}{3}+\frac{2}{11})}=\frac{55}{118}$
$\Rightarrow P\left(E_3/E\right)=\frac{\frac{1}{3}\left(\frac{2}{11}\right)}{\frac{1}{3}(\frac{1}{5}+\frac{1}{3}+\frac{2}{11})}=\frac{30}{118}$