1. 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.
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By: anil on 05 May 2019 01.39 am
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 = $$frac{4}{10} imes frac{3}{7}$$ = $$frac{6}{35}$$
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