Question 47
5 red, 4 blue and 3 green balls are kept in bag. Two balls are drawn at random from the bag. What is the probability that none of the two balls drawn is red in color?
Solution
Probability of 1st ball being Non-red => $$ \frac{7}{12} $$
Probability of 2nd ball being Non-red => $$ \frac{6}{11} $$
Since, both the balls being non-red => $$ \frac{6}{11} \times \frac{7}{12} $$ => $$ \frac{7}{22} $$
Option D is correct
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