Question 112

A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag, and the probability that both are green.

Solution

Total balls in bag A = 4 + 6 = 10

Probability that ball is green = $$\frac{4}{10}$$

Total balls in bag B = 3 + 4 = 7

Probability that ball is green = $$\frac{3}{7}$$

=> Required probability = $$\frac{4}{10} \times \frac{3}{7}$$

= $$\frac{6}{35}$$

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: