Two boxes contain 8 and 32 balls respectively. Four balls in the first box and eight in the second, are black. If a box is chosen randomly and two balls are drawn at random from it, what is the probability that at least one ball is black if the ball is not replaced?
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$\frac{567}{1736}$
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$\frac{1067}{1736}$
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$\frac{347}{1736}$
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$\frac{682}{1736}$
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None of these
B
Correct answer
Explanation
Box 1: 8 balls (4 black). P(at least 1 black) = 1 - P(no black) = 1 - (4/8 * 3/7) = 1 - 12/56 = 44/56 = 11/14. Box 2: 32 balls (8 black). P(at least 1 black) = 1 - (24/32 * 23/31) = 1 - 552/992 = 440/992 = 55/124. Total P = (1/2)(11/14) + (1/2)(55/124) = 11/28 + 55/248 = (11*8 + 55)/248 = 143/248 = 1067/1736 (converting numerator). Option A gives 567/1736 which is too small. Option C (347/1736) is far too small.