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better chance of winning

by abhi332 » Thu Feb 25, 2010 1:15 pm
Mike recently won a contest in which he will have the opportunity to shoot free throws in order to win $10,000. In
order to win the money Mike can either shoot 1 free throw and make it, or shoot 3 free throws and make at least
2 of them. Mike occasionally makes shots and occasionally misses shots. He knows that his probability of making
a single free throw is p, and that this probability doesn't change. Would Mike have a better chance of winning if
he chose to attempt 3 free throws?
(1) p < 0.7
(2) p > 0.6
[spoiler]
OA:B[/spoiler]
What you think, you become.
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Source: — Data Sufficiency |
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by hrishi19884 » Fri Feb 26, 2010 1:14 am
Was a diff question for me ;)


https://gmatclub.com:8080/forum/viewtopic.php?p=652513
Hrishi

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by Fiver » Sat Feb 27, 2010 6:57 am
abhi332 wrote:Mike recently won a contest in which he will have the opportunity to shoot free throws in order to win $10,000. In
order to win the money Mike can either shoot 1 free throw and make it, or shoot 3 free throws and make at least
2 of them. Mike occasionally makes shots and occasionally misses shots. He knows that his probability of making
a single free throw is p, and that this probability doesn't change. Would Mike have a better chance of winning if
he chose to attempt 3 free throws?
(1) p < 0.7
(2) p > 0.6
[spoiler]
OA:B[/spoiler][/quote
P ( winnning in 3 attempts) :
P ( hitting target in all 3 attempts) + P (hitting target in atleast 2)

= p^3 + 3 (p^2 (1- p ))
= p^3 + 3p^2 - 3p^3
= -2p^3 + 3p^2

Rephrase question is p> -2p^3 + 3p^2
after simpifying we get :

is 2p^2 - 3p + p > 0 ?
boils down to is p> 1/2?

Hence B
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