Question

Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls. Out of these two boxes one is selected at random and the probability of choosing box A is double that of box B. If a red ball is drawn from the selected box, then the probability that it has come from box B is

• A. $\frac{21}{41}$
• B. $\frac{10}{31}$
• C. $\frac{12}{31}$
• D. $\frac{13}{41}$

Answer 1

The correct option is B

$\frac{10}{31}$

Explanation for correct options:

Considering elements of Box A and Box B:

Number of red balls in Box A = 3

Total number of balls = 5

Probability of selecting red balls from Box A = $\frac{3}{5}$

Let, Probability of selecting Box A be 2a

Number of red balls in Box B = 4

Total number of balls = 7

Probability of selecting red balls from Box B = $\frac{4}{7}$

Let, Probability of selecting Box B be a

Probability of drawing a red ball from Box B = $\frac{a({\displaystyle \frac{4}{7}})}{a({\displaystyle \frac{4}{7}})+2a({\displaystyle \frac{3}{5}})}$

= $\frac{{\displaystyle \frac{4}{7}}}{{\displaystyle \frac{4}{7}}+2({\displaystyle \frac{3}{5}})}$

= $\frac{{\displaystyle \frac{4}{7}}}{{\displaystyle \frac{4}{7}}+{\displaystyle \frac{6}{5}}}$

= $\frac{20}{62}$

= $\frac{10}{31}$

Therefore, the probability of drawing a red ball from the Box B is $\frac{10}{31}$

Hence, Option (B) is correct.