Data Sufficiency

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.


The answer is D.

Can someone explain to me how to even begin to evaluate these statements?

Thank you

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.
Statement 1 implies the following two facts:
Fact 1: The die is thrown two times (implying that ONE MORE TIME = THREE throws).
Fact 2: P(x) = 1/5.
Since the two statements cannot contradict each other, these two facts must also satisfy statement 2.

TEST these two facts in 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.

Alternate way to evaluate statement 2:

Statement 2: If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.
Thus, P(no x's) = 1 - 61/125 = 64/125.

Since the die has x sides, P(not x) = (x-1)/x.
Let n = the number of throws if the die is thrown one more time.
Thus:
((x-1)/x)^n = 64/125.

Since 64 = 4*4*4 and 125 = 5*5*5, the only combination that works is x=5 and n=3:
((5-1)/5)³ = 64/125.
(4/5)³ = 64/125.

Thus, P(not x) = 4/5.
Since n=3 is the number of throws if the die is tossed one more time, the actual number of throws is 2.
In two throws, P(at least one x) = 1 - P(no x's) = 1 - (4/5)² = 9/25.
SUFFICIENT.

Can you please go into a bit more detail about why you can assum 3 throws? It seems confusing to me like you are using the information from Statement 1. I know you aren't, just don't understand the logic.