If one bucket contains 8 red balls and 2 black balls and another bucket contains 7 red balls and 3 black balls, the probability of having at least one red ball from drawings of one ball from each of the two buckets is
Options:
(A) 0.94
(B) 0.84
(C) 0.56
(D) 0.38
Official Answer Key:
A
Step-by-Step Solution & Explanation
P(red from bucket 1) = 8/10 = 0.8; P(red from bucket 2) = 7/10 = 0.7. P(at least one red) = 1 − P(no red from either) = 1 − (1−0.8)(1−0.7) = 1 − 0.2×0.3 = 1 − 0.06 = 0.94.
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