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 = $$\frac{4}{10} \times \frac{3}{7}$$
= $$\frac{6}{35}$$
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