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Updated on: 19 Sep 2013, 05:09
Originally posted by ctrlaltdel on 15 Nov 2009, 22:21.
Last edited by Bunuel on 19 Sep 2013, 05:09, edited 1 time in total.
Last edited by Bunuel on 19 Sep 2013, 05:09, edited 1 time in total.
Added the OA.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
49% (02:31) correct
51%
(02:46)
wrong
based on 794
sessions
History
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Time
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Not Attempted Yet
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
(1) p < 0.7
(2) p > 0.6
16 Apr 2014, 02:26
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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?
The probability of winning for 1 throw = p;
The probability of winning for 3 throw = \(C^2_3p^2(1-p)+p^3=3p^3(1-p)+p^3=p^2(3-2p)\).
The question asks: is \(p^2(3-2p)>p\)? --> is \(p(3-2p)>1\)? --> is \(2p^2-3p+1<0\)? is \((p-\frac{1}{2})(p-1)<0\)? --> is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B.
The probability of winning for 1 throw = p;
The probability of winning for 3 throw = \(C^2_3p^2(1-p)+p^3=3p^3(1-p)+p^3=p^2(3-2p)\).
The question asks: is \(p^2(3-2p)>p\)? --> is \(p(3-2p)>1\)? --> is \(2p^2-3p+1<0\)? is \((p-\frac{1}{2})(p-1)<0\)? --> is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B.
16 Nov 2009, 01:19
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Good problem...
Statement 1:
Probability of a single free thow: Less than 0.7
Probability of making at least two out of three:
Probability of making two out of three: (.7)*(.7)*(.3)
Probability of making three out of three (.7)*(.7)*(.7)
There are three ways of making two out of three so the total probability of making at least two out of three is:
(.7)*(.7)*(.3)*3+(.7)*(.7)*(.7) = .784
OR
Less than .784
In this case, going for the three free throws is better...
But if the probability of making the shot was .1:
(.1)*(.1)*(.9)*3+(.1)*(.1)*(.1) = .028
going for a single free throw is better:
INSUFFICIENT
Statement 2:
Probability of a single free thow: Greater than 0.6
Probability of making at least two out of three:
(.6)*(.6)*(.4)*3+(.6)*(.6)*(.6) = .648
Going for the three free throws is better
If the probability of making the shot was .9:
(.9)*(.9)*(.1)*3+(.9)*(.9)*(.9) = .972
going for three free throws is better!
SUFFICIENT
ANSWER: B.
What is the OA? Does anyone have a faster way to do this? If I ran into this problem on the GMAT, this would have taken me 5+ mins to do...
Statement 1:
Probability of a single free thow: Less than 0.7
Probability of making at least two out of three:
Probability of making two out of three: (.7)*(.7)*(.3)
Probability of making three out of three (.7)*(.7)*(.7)
There are three ways of making two out of three so the total probability of making at least two out of three is:
(.7)*(.7)*(.3)*3+(.7)*(.7)*(.7) = .784
OR
Less than .784
In this case, going for the three free throws is better...
But if the probability of making the shot was .1:
(.1)*(.1)*(.9)*3+(.1)*(.1)*(.1) = .028
going for a single free throw is better:
INSUFFICIENT
Statement 2:
Probability of a single free thow: Greater than 0.6
Probability of making at least two out of three:
(.6)*(.6)*(.4)*3+(.6)*(.6)*(.6) = .648
Going for the three free throws is better
If the probability of making the shot was .9:
(.9)*(.9)*(.1)*3+(.9)*(.9)*(.9) = .972
going for three free throws is better!
SUFFICIENT
ANSWER: B.
What is the OA? Does anyone have a faster way to do this? If I ran into this problem on the GMAT, this would have taken me 5+ mins to do...
