BTGmoderatorDC wrote:Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?
(1) The probability of selecting exactly 2 apples is greater than 1/2.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.
OA C
Source: Manhattan Prep
Say there are x numbers of apples, thus, there are (8 - x) numbers of bananas.
Let's take each statement one by one.
(1) The probability of selecting exactly 2 apples is greater than 1/2.
=> xC2 / 8C2 > 1/2
[x*(x - 1) / 8.7] > 1/2 => x(x - 1) > 28
For the inequality to hold true, x = 6, 7 or 8. Note that x ≠8, else there would not be any banana left. Thus, x = 6 or 7. Thus, the numbers of bananas = 1 or 2.
Case 1: If numbers of bananas = 1, the probability of selecting exactly 2 bananas = 0 (an impossible even)
Case 2: If numbers of bananas = 2, the probability of selecting exactly 2 bananas = 2C2/8C2 = 1/8C2 = 1/28
No unique answer. Insufficient.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.
=> [x.(8 - x)] / 8C2 = [2.x.(8 - x)] / [8.7] > 1/3 => x .(8 - x) > 9.33 => x = 6/5/4/3/2
No unique answer. Insufficient.
(1) and (2) together
The only common value is x = 6. Thus, the probability of selecting exactly 2 bananas = 2C2/8C2 = 1/8C2 = 1/28. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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