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P(100) = \(\frac{1}{10}\)

P(not 100) = \(\frac{9}{10}\)

P(200) = \(\frac{1}{8}\)

P(not 200) = \(\frac{7}{8}\)

Independent events: Only one: P(100) * P(not 200) + p(not 100) * P(200)

=> \(\frac{1}{10} * \frac{7}{8} + \frac{9}{10} * \frac{1}{8}\)

=> \(\frac{7}{80} + \frac{9}{80}\)

=> \(\frac{16}{80} = \frac{1}{5}\)

Answer B
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Bunuel
At a carnival, the probability that a boy will win the 100 m event is 1/10, and the probability he wins the 200 m event is 1/8. If these events are independent, what is the probability that he wins only one of these events?

A. 1/80
B. 1/5
C. 17/80
D. 9/40
E. 4/5



hi KarishmaB

when we get 1/10 * 7/8= 7/80 why we dont result multiply by 2!

I mean that 100m win and 200 m lose is one event and 200m lose and 100m win is another event

same question applies to 1/8*9/10

For example in this question we consider different arrangements https://gmatclub.com/forum/there-are-10 ... l#p3203227
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dave13
Bunuel
At a carnival, the probability that a boy will win the 100 m event is 1/10, and the probability he wins the 200 m event is 1/8. If these events are independent, what is the probability that he wins only one of these events?

A. 1/80
B. 1/5
C. 17/80
D. 9/40
E. 4/5



hi KarishmaB

when we get 1/10 * 7/8= 7/80 why we dont result multiply by 2!

I mean that 100m win and 200 m lose is one event and 200m lose and 100m win is another event

same question applies to 1/8*9/10

For example in this question we consider different arrangements https://gmatclub.com/forum/there-are-10 ... l#p3203227

Whether to multiply the probability by 2 or not (arrange or not) is one of the hardest things to explain.

I mean that 100m win and 200 m lose is one event and 200m lose and 100m win is another event

They are not two different cases. Whether you win the 100 m first or lose the 200 m first is irrelevant. The overall result is that you won the 100 m and lost the 200 m. That is the result. You ran the 100m when it was scheduled and ran the 200 m when it was scheduled. You did not choose the sequence and the sequence does not change the overall result. The sequence is pre-defined.
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