Question

There are two bags $A$ and $B$. Bag A contains $3$ white and $4$ red balls whereas bag $B$ contains $4$ white and $3$ red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probability that these were drawn from bag $B$.

• A. $\frac{7}{11}$
• B. $\frac{5}{11}$
• C. $\frac{3}{11}$
• D. $\frac{9}{11}$

Answer 1

The correct option is D $\frac{9}{11}$

$E_1=BagA$ $E_2=BagB$

$P\left(E_1\right)=\frac{1}{2}=P\left(E_2\right)$

Probability of drawing 2 white and 2 red from Bag A

$P\left(\frac{A}{E_1}\right)=\frac{3_{C_2}\times 4_{C_1}}{7_{C_3}}=\frac{12}{35}$

Probability of drawing 2 white and 1 red ball from Bag B

$P\left(\frac{A}{E_2}\right)=\frac{4_{C_2}\times 3_{C_1}}{7_{C_3}}=\frac{4\times 3}{2}\times \frac{3\times 1}{7\times 6\times 5}\times 3\times 2=\frac{18}{35}$

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

$=\frac{\frac{1}{2}\times \frac{18}{35}}{\frac{1}{2}\times \frac{12}{35}+\frac{1}{2}\times \frac{18}{35}}$

$=\frac{\frac{18}{70}}{\frac{12}{70}+\frac{18}{70}}$

$=\frac{18}{30}$

$=\frac{9}{15}$