tpz wrote:Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?
(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.
P(at least one x) = 1 - P(no x's).
Statement 1: The probability of getting an x on either of THE TWO TIMES that Rachel throws the die is 1/5.
The die is being thrown TWO TIMES.
Since P(x) = 1/5, P(not x) = 4/5.
Thus, P(no x's) = 4/5 * 4/5 = 16/25.
Thus, P(at least one x) = 1 - 16/25 = 9/25.
SUFFICIENT.
Statement 2: If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.
For the correct answer to be D, statement 2 must offer the SAME TWO FACTS as statement 1:
Fact 1: The die is thrown two times (implying that ONE MORE TIME = THREE throws).
Fact 2: P(x) = 1/5.
Check whether these two facts satisfy statement 2:
P(not X on each of 3 throws) = 4/5 * 4/5 * 4/5 = 64/125.
P(at least one x) = 1 - 64/125 = 61/125.
Success!,
Since P(x) = 1/5 and a total three throws constitute the ONLY combination that will yield a result of 61/125, we know that statement 2 offers the SAME TWO FACTS as statement 1, implying that Rachel actually throws the die TWO TIMES and that P(at least one x) = 9/25.
SUFFICIENT.
The correct answer is
D.
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