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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%
(02:47)
wrong
based on 805
sessions
History
Date
Time
Result
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
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) p < 0.7
(2) p > 0.6
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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...
