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John and Peter
compete against each other in shooting. Proba [#permalink]
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John and Peter compete against each other in shooting. Probability of John hitting the target is 0.4 and Probability of Peter hitting the target is 0.7. If John has the first shot, post which they strike alternately and game ends when someone hits the target first. What is the probability that John
wins?
A. 8/13
B. 15/22
C. 20/41
D. 2/5
E. 3/25
Posted from my mobile device
Math Expert
Joined: 02 Aug 2009
Posts: 10596
John and Peter compete against each other in shooting. Proba
[#permalink]
Dillesh4096 wrote:
John and Peter compete against each other in shooting. Probability of John hitting the target is 0.4 and
Probability of Peter hitting the target is 0.7. If John has the first shot, post which they strike alternately and game ends when someone hits the target first. What is the probability that John wins?
A. 8/13
B. 15/22
C. 20/41
D. 2/5
E. 3/25
Posted from my mobile device
John can win when he hits the target but Peter misses it in previous attempts..
John can win in first shot of the game =0.4
or third shot of the
game, when peter misses the second one = [1-0.4][1-0.7]0.4 =0.6*0.7*0.4 and so on...
Over all probability = \[0.4+0.6*0.3*0.4+0.6*0.3*0.6*0.3*0.4...=0.4[1+0.6*0.3+[0.6*0.3]^2+....]\]
So what we have in the bracket is a geometrical progression with each term increasing by 0.6*0.3 over the preceding term
Sum = \[0.4[\frac{a}{1-r}]=0.4[\frac{1}{1-0.18}]=0.4*\frac{1}{0.82}=\frac{40}{82}=\frac{20}{41}\]
C
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Joined: 01 Jun 2019
Posts: 78
Location: United States
Concentration: Strategy
GMAT 1: 740 Q49 V41
Re: John and Peter compete against each other
in shooting. Proba [#permalink]
The question statement is ambiguous.
Re: John and Peter compete against each other in shooting. Proba [#permalink]
30 Oct 2019, 08:04
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