Question

An urn $A$ contain $8$ black balls and $5$ white balls. A second urn $B$ contains $6$ black and $7$ white balls. The probability that blind folded person in one draw shall obtain a white ball is

• A. $\frac{5}{13}$
• B. $\frac{7}{13}$
• C. $\frac{6}{13}$
• D. $\frac{5}{26}$

Answer 1

The correct option is C $\frac{6}{13}$

Probability of White Balls in Urn A $=\frac{WhiteBallsofUrnA}{TotalBallsofUrnA}$

Probability of White Balls in Urn A $=\frac{5}{13}$

Probability of White Balls in Urn B $=\frac{WhiteBallsofUrnB}{TotalBallsofUrnB}$

Probability of White Balls in Urn B $=\frac{7}{13}$

Probability of White Balls in Urn A and Urn B $=\frac{1}{2}[\frac{5}{13}+\frac{7}{13}]$

$\to \frac{6}{13}$

Thus the correct answer is $C$