AAPL wrote:Manhattan Prep
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 \(1/3\).
What is the value of P(BB)?
Statement 1:
From 8 pieces of fruit, the probability of picking an apple = A/8.
After 1 of the A apples has been picked, the number of remaining apples = A-1.
From the 7 remaining pieces of fruit, the probability of picking one of the A-1 remaining apples = (A-1)/7.
When we want events happening together, we MULTIPLY the probabilities:
A/8 * (A-1)/7
Since the resulting is greater than 1/2, we get:
A/8 * (A-1)/7 > 1/2
(A² - 1)/56 > 1/2
A² > 28
Integer options such that A² > 28 and A+B=8:
A=6, B=2
A=7, B=1
A=8, B=0
Since the three cases will yield different values for P(BB), INSUFFICIENT.
Statement 2:
From 8 pieces of fruit, the probability of picking an apple = A/8.
From the 7 remaining pieces of fruit, the probability of picking a banana = B/7.
When we want events happening together, we MULTIPLY the probabilities:
A/8 * B/7
Since the apple and banana can be picked in either order -- AB or BA -- we multiply by 2:
A/8 * B/7 * 2
Since the resulting probability is greater than 1/3, we get:
A/8 * B/7 * 2 > 1/3
(AB)/28 > 1/3
AB > 28/3
AB > 9.33
Integer options such that AB>9.33 and A+B=8:
A=2, B=6
A=3, B=7
A=4, B=4
A=5, B=3
A=6, B=2
Since the various cases will yield different values for P(BB), INSUFFICIENT.
Statements combined:
Only the case in blue satisfies both statements:
A=6, B=2
Thus, the value of P(BB) can be determined.
SUFFICIENT.
The correct answer is
C.
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