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.
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