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Mike recently won a contest in which he will have the opport
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Updated on: 19 Sep 2013, 05:09
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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? (1) p < 0.7 (2) p > 0.6
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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.
Added the OA.




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Re: Mike recently won a contest in which he will have the opport
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16 Apr 2014, 02:26
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 = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\). The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)? (1) p < 0.7. Not sufficient. (2) p > 0.6. Sufficient. Answer; B.
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Re: Mike Shooting free throws
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16 Nov 2009, 01:19
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...




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Re: Mike recently won a contest in which he will have the opport
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24 Feb 2014, 14:15
ctrlaltdel 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 I didn't quite get this one This is my procedure OK if he shot one time only probability will be 'p' If he makes at least two shots then probability will be: Probability of making all three shots + Probability of making 2 shots and missing 1 shot Let's see probability of making all three is p^3 Probability of making 2 out of 3 is 3C2 * p^2* (1p) So question goes down to is p> 3(p^2)(1p) + p^3? Is 2p^2  3p + 1 <0? Statement 1 If we replace 0.7 for p we get that yes, it is always negative hence true Sufficient Statement 2 Clearly insufficient. Could someone explain if I'm wrong somewhere? I'm mainly concerned about the formulation of the question rather than in the solving part but still could have made an error Many thanks Cheers J



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Re: Mike recently won a contest in which he will have the opport
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24 Feb 2014, 21:37
jlgdr wrote: ctrlaltdel 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 I didn't quite get this one This is my procedure OK if he shot one time only probability will be 'p' If he makes at least two shots then probability will be: Probability of making all three shots + Probability of making 2 shots and missing 1 shot Let's see probability of making all three is p^3 Probability of making 2 out of 3 is 3C2 * p^2* (1p) So question goes down to is p> 3(p^2)(1p) + p^3? Is 2p^2  3p + 1 <0? Statement 1 If we replace 0.7 for p we get that yes, it is always negative hence true Sufficient Statement 2 Clearly insufficient. Could someone explain if I'm wrong somewhere? I'm mainly concerned about the formulation of the question rather than in the solving part but still could have made an error Many thanks Cheers J First of all, my friend, I think you have made a mistake in opening the brackets. I got it as 3p^2 2p^3 p < 0 (I'm bad at algebra ) Secondly, Why to complicate this question? By reading it you can understand it will involve cubes and squares + fractions. So its rather better to put your own values (under the permissible limit). My friend, solve question strategically and not mathematically Third, this may not sound the best way to solve but you can do a little thinking before diving in with values. A probability of < 0.7 can mean it could 0.1 or 0.6 if its 0.1 > if you are very lucky you might get a throw in once chance but 2 out 3 throws could mean tough for you. if its 0.6 > More than half of the time you get a shot. You might or might not get it in one chance but you are better off with taking three shots as you are aware half of the times you get the shot INSUFFICIENT p > 0.6 Obviously here you can easily say, your probability is quite good. Although you don't get every shot (p=1, read question he occasionally misses the shot so p =1 is not possible ) but you mostly get the shot so you better go for three shots than one. It didn't take me much time to think this obviously reading/writing this might seems like a long process but it seriously doesn't even take more than half a minute. After this, you can assume some value to confirm SUFFICIENT Note that had they not written "occasionally he misses the shot" it would have been tough as p> 0.6 could mean p=1 So if you always get a shot than taking one or three doesn't matter.



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Re: Mike recently won a contest in which he will have the opport
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15 Apr 2014, 17:42
Thanks Buddy, agree that algebraic approach to this question could prove painful? Bunuel, do you want to take a stab at this one and show us how you roll? Thanks Cheers! J



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Re: Mike recently won a contest in which he will have the opport
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16 Apr 2014, 05:50
Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. could you please explain how you got this? \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).



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Re: Mike recently won a contest in which he will have the opport
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18 Apr 2014, 10:55
borntobreaktherecord wrote: Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. could you please explain how you got this? \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\). At least 2 hits in 3 shots means 2 or 3 hits: The probability of 2 hits in 3 shots is \(P(HHN) = C^2_3p^2(1p)\), we are multiplying by \(C^2_3=3\) because HHN scenario can occur in 3 ways: HHN, HNH, NHH. The probability of 3 hits in 3 shots is \(P(HHH) = p^3\). Thus the overall probability for at least 2 hits in 3 shots is \(C^2_3p^2(1p)+p^3\). Hope it's clear.
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Re: Mike recently won a contest in which he will have the opport
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23 Apr 2014, 08:42
Hi Bunuel,
why do we need to consider the case HHN , i mean once he gets two hits ......wouldnt that suffice??



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Re: Mike recently won a contest in which he will have the opport
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10 May 2014, 10:38
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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Re: Mike recently won a contest in which he will have the opport
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10 Jun 2014, 10:29
anjubaid wrote: Hi Bunuel,
why do we need to consider the case HHN , i mean once he gets two hits ......wouldnt that suffice?? Yes, you're right here. We get the same answer either way. Considering just the option 'HH' includes both 'HHM' and 'HHH'; Bunuel just counted them separately in his answer. Notice that the probability of HHM + probability of HHH = probability of HH: \(p^2(1p) + p^3 = p^2  p^3 + p^3 = p^2.\) Careful though! If you just consider HH and NOT HHM, then DO NOT include HHH, as it's already included in HH. So yes, you're correct; it's just a matter of which approach you like best.



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Re: Mike recently won a contest in which he will have the opport
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09 Oct 2014, 08:08
Bunuel wrote: borntobreaktherecord wrote: Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. could you please explain how you got this? \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\). At least 2 hits in 3 shots means 2 or 3 hits: The probability of 2 hits in 3 shots is \(P(HHN) = C^2_3p^2(1p)\), we are multiplying by \(C^2_3=3\) because HHN scenario can occur in 3 ways: HHN, HNH, NHH. The probability of 3 hits in 3 shots is \(P(HHH) = p^3\). Thus the overall probability for at least 2 hits in 3 shots is \(C^2_3p^2(1p)+p^3\). Hope it's clear. Can you please clarify why order does matter here??I solved this problem after solving a problem similar to it,the one where triplets Adam,Bruce and Charlie enter a triathlon.There we needed to find probability that atleast 2 triplets will win a medal.Here we have to find probability of at least two throws going correctly. But still in this problem we are considering order,but not in the triplets problem. Please clarify.



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Re: Mike recently won a contest in which he will have the opport
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09 Oct 2014, 08:28
aakafh wrote: Bunuel wrote: borntobreaktherecord wrote:
could you please explain how you got this? \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
At least 2 hits in 3 shots means 2 or 3 hits: The probability of 2 hits in 3 shots is \(P(HHN) = C^2_3p^2(1p)\), we are multiplying by \(C^2_3=3\) because HHN scenario can occur in 3 ways: HHN, HNH, NHH. The probability of 3 hits in 3 shots is \(P(HHH) = p^3\). Thus the overall probability for at least 2 hits in 3 shots is \(C^2_3p^2(1p)+p^3\). Hope it's clear. Can you please clarify why order does matter here??I solved this problem after solving a problem similar to it,the one where triplets Adam,Bruce and Charlie enter a triathlon.There we needed to find probability that atleast 2 triplets will win a medal.Here we have to find probability of at least two throws going correctly. But still in this problem we are considering order,but not in the triplets problem. Please clarify. Forget at least 2. We are talking now about 2 hits in 3 throws: HHN. We can have 2 hits in 3 throws in following ways: HHN, HNH, NHH. Each of which has its probability. Sorry, cannot explain any better.
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Mike recently won a contest in which he will have the opport
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25 Aug 2015, 14:02
Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. Awesome. Good to know that my method of solving it matched with yours. Do we have some more problems in which we can apply same binomial distribution concept? Thanks, Gaurav



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Re: Mike recently won a contest in which he will have the opport
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25 Aug 2015, 23:41
GauravSolanky wrote: Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. Awesome. Good to know that my method of solving it matched with yours. Do we have some more problems in which we can apply same binomial distribution concept? Thanks, Gaurav Some questions to practice: whatistheprobabilitythatafieldgunwillhitthriceon127334.htmlthereisa90chancethataregisteredvoterinburghtown56812.htmlcombinationps55071.htmltheprobabilitythatafamilywith6childrenhasexactly88945.html? Hope it helps.
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Re: Mike recently won a contest in which he will have the opport
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26 Aug 2015, 09:57
One thing to clarify, 3 free shots in the question means  one shot after another shot and then one more shot, am I right?



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Re: Mike recently won a contest in which he will have the opport
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27 Aug 2015, 01:06
wintersrhn wrote: One thing to clarify, 3 free shots in the question means  one shot after another shot and then one more shot, am I right? __________ Yes. How else?
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Re: Mike recently won a contest in which he will have the opport
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16 Nov 2016, 08:12
Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. Hi Bunuel, Firstly, thank you for all the work you have done on this site. my doubt is why are we calculating all three wins i.e p^3; shouldn't we just write p^2 for two consec wins as the third win wont matter?



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Re: Mike recently won a contest in which he will have the opport
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16 Nov 2016, 08:16
shivamtt770 wrote: Bunuel 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?
The probability of winning for 1 throw = p; The probability of winning for 3 throw = \(C^2_3p^2(1p)+p^3=3p^3(1p)+p^3=p^2(32p)\).
The question asks: is \(p^2(32p)>p\)? > is \(p(32p)>1\)? > is \(2p^23p+1<0\)? is \((p\frac{1}{2})(p1)<0\)? > is \(\frac{1}{2}<p<1\)?
(1) p < 0.7. Not sufficient.
(2) p > 0.6. Sufficient.
Answer; B. Hi Bunuel, Firstly, thank you for all the work you have done on this site. my doubt is why are we calculating all three wins i.e p^3; shouldn't we just write p^2 for two consec wins as the third win wont matter? Check here: mikerecentlywonacontestinwhichhewillhavetheopport86901.html#p1357313
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Re: Mike recently won a contest in which he will have the opport
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07 Jan 2017, 06:39
The question requires us to determine whether Mike's odds of winning are better if he attempts 3 shots instead of 1. For that to be true, his odds of making 2 out of 3 must be better than his odds of making 1 out of 1. There are two ways for Mike to at least 2 shots: Either he hits 2 and misses 1, or he hits all 3: In order for this inequality to be true, p must be greater than .5 but less than 1 (since this is the only way to ensure that the left side of the equation is negative). But we already know that p is less than 1 (since Mike occasionally misses some shots). Therefore, we need to know whether p is greater than .5. If it is, then the inequality will be true, which means that Mike will have a better chance of winning if he takes 3 shots. Statement 1 tells us that p < .7. This does not help us to determine whether p > .5, so statement 1 is not sufficient. Statement 2 tells us that p > .6. This means that p must be greater than .5. This is sufficient to answer the question. The correct answer is B: Statement (2) alone is sufficient, but statement (1) alone is not sufficient. Refer attachment to see the calculation.
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prob.PNG [ 19.06 KiB  Viewed 3622 times ]
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