both Peter and Paul win

sanju09 wrote:The probability that Peter would win a grill show is twice of the probability that Paul would win the same grill show. If the probability that both Peter and Paul win the grill show is 9/32, what is the probability that only one of the two would win the grill show?
A. ¾
B. 9/16
C. 3/8
D. ¼
E. 5/32

Let P(Paul wins) = x.
Since Peter's probability is twice Paul's, P(Peter wins) = 2x.
Since the probability that both win is 9/32, we get:
x * 2x = 9/32.
2x² = 9/32
x² = 9/64
x = 3/8.

Thus:
P(Paul wins) = x = 3/8, implying that P(Paul loses) = 5/8.
P(Peter wins) = 2x = 2 * (3/8) = 3/4, implying that P(Peter loses) = 1/4.

Case 1: P(Paul wins but Peter doesn't) = 3/8 * 1/4 = 3/32.
Case 2: P(Peter wins but Paul doesn't) = 3/4 * 5/8 = 15/32.
Since exactly one person wins in either Case 1 or Case 2, we ADD the fractions:
3/32 + 15/32 = 18/32 = 9/16.

The correct answer is B.

You might want to stipulate that the two men's chances of winning are independent, as the GMAC has been dabbling with dependent probability of late, at least going by the Question Pack they just released. (Probability in general seems to be a bigger focus of the test, though the new OG will tell us more.)

Anyway, I thought of this problem as:

(Prob ONLY One Wins) = 1 - (Both Win) - (Neither Wins)
= 1 - (x*2x) - (1-x)(1-2x)
= 3x - 4x²
= x * (3 - 4x)

We're given that 2x² = 9/32, or x = 3/8, so we have

(3/8) * (3 - 3/2), or (3/8) * (3/2), or 9/16.

Great replies!

If P(Paul wins) = x then P(Peter wins) = 2x; such that (x)(2x) = 9/32 or x = 3/8 (probability that Paul loses is 5/8) and 2x = ¾ (probability that Peter loses is ¼).

Now, the probability that only one of the two would win the grill show means

Peter wins and Paul loses OR Paul wins and Peter loses

= (¾) (5/8) + (3/8)( ¼)

= (15/32) + (3/32)

[spoiler]= 18/32

= 9/16.

Chose (B).[/spoiler]