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At a carnival, the probability that a boy will win the 100 m event is

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Bunuel
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At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   01 Mar 2021, 03:45
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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
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B
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Re: At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   01 Mar 2021, 06:17
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P(Winning 100 m) = 1/10. Therefore P(Losing the 100m) = 1 - 1/10 = 9/10

P(Winning 200 m) = 1/8. Therefore P(Losing the 100m) = 1 - 1/8 = 7/8

P(Winning exactly 1 event) = Winning 100m and losing 200 m or Losing 100m and winning 200m

= (1/10 * 7/8) + (9/10 * 1/8) = 7/80 + 9/80 = 1680 = 1/5


Option B

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Re: At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   01 Mar 2021, 08:15
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Bunuel wrote:
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


In order to win one event ONLY, he needs to win either one then lose the other one.

\(\frac{1}{10}*\frac{7}{8} + \frac{1}{8}*\frac{9}{10} = \frac{7 + 9}{80} = \frac{1}{5}\)

Ans: B
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Re: At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   01 Mar 2021, 08:45
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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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At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   20 May 2023, 00:03
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Bunuel wrote:
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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Re: At a carnival, the probability that a boy will win the 100 m event is [#permalink] New post   21 May 2023, 21:14
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dave13 wrote:
Bunuel wrote:
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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Re: At a carnival, the probability that a boy will win the 100 m event is [#permalink]
21 May 2023, 21:14
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