Question

An urn contains $5$ white and $7$ black balls. A second urn contains $7$ white and $8$ black balls. One ball is transfered from the $1^{st}$ urn to the $2^{nd}$ urn without noticing its colour. A ball is now drawn at random from the $2^{nd}$ urn. The probability that it is white is

• A. $\frac{49}{192}$
• B. $\frac{39}{192}$
• C. $\frac{99}{192}$
• D. $\frac{89}{192}$

Answer 1

The correct option is D $\frac{89}{192}$

Let $E_1$ be the event first ball is white and $E_2$ be the event first ball is black.
$P\left(E_1\right)=\frac{5}{12},P\left(E_2\right)=\frac{7}{12}$
Let $E$ be the event ball drawn from $2^{nd}$ urn is white.
$P\left(\frac{E}{E_1}\right)=\frac{8}{16}$ and$P\left(\frac{E}{E_2}\right)=\frac{7}{16}$
Using Total Theorem of probablity,
$P\left(E\right)=P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)$
$P\left(E\right)=\frac{5}{12}\times \frac{8}{16}+\frac{7}{12}\times \frac{7}{16}=\frac{40+49}{192}=\frac{89}{192}$