A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.
A. 12/91 B. C. D.
A. 12/91 B. C. D.
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Correct Answer: Option
Explanation:
Total balls = 8+4+2= 14
n(R) = 8, n (B) =4, n(G) = 2
Without replacement, it is p(RB) or p(BR)
= (8/14 x 4/13) + (4/14 + 8/13) = 16/91 + 16/91
= 32/91
Total balls = 8+4+2= 14
n(R) = 8, n (B) =4, n(G) = 2
Without replacement, it is p(RB) or p(BR)
= (8/14 x 4/13) + (4/14 + 8/13) = 16/91 + 16/91
= 32/91