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805+ (Hard)|   Probability|      
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gmatkiller666
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Mike recently won a contest in which he will have the opport 28 Jan 2017, 21:33
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anairamitch1804
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.
Show more


Hello, My inequality skills are a bit short.

When you get to the last line of the IE equation

How do we get from (p-.5)(p-1) < 0

to 1/2 < p < 1 ?
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Mike recently won a contest in which he will have the opport 29 Jan 2017, 03:47
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gmatkiller666
anairamitch1804
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.


Hello, My inequality skills are a bit short.

When you get to the last line of the IE equation

How do we get from (p-.5)(p-1) < 0

to 1/2 < p < 1 ?
Show more

(p - 1/2)(p - 1) < 0.

The roots are 1/2 and 1. "<" sign indicates that the solution lies between the roots, thus 1/2 < p < 1.

Check for more the links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

https://gmatclub.com/forum/inequalities-trick-91482.html
https://gmatclub.com/forum/data-suff-ine ... 09078.html
https://gmatclub.com/forum/range-for-var ... 09468.html
https://gmatclub.com/forum/everything-is ... 08884.html
https://gmatclub.com/forum/graphic-appro ... 68037.html

Hope this helps.
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Mike recently won a contest in which he will have the opport 27 Mar 2017, 13:34
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What if the probability of making a successful free throw is 1, in that case option B, p>0.6 cannot be the answer, since the question is really asking if 0.5<p<1 ?
Please correct me if my understanding is wrong..
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Mike recently won a contest in which he will have the opport 27 Jul 2017, 00:41
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aakash1618

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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Another way to arrive at this is just to work out the cases. If he has 3 shots and must make two, then he can win if:

HH (He makes the first two shots. Third shot is not needed, he already won.)
NHH (Missed the first, made the second two)
HNH (Made the first and the last)

the probabilities of these three winning scenarios in terms of "p" are:

HH ---> \(p*p\) --------------------> \(p^2\)
NHH ---> \((1-p)*p*p\)--------------> \(p^2(1-p)\)
HNH----> \((1-p)*p*p\)--------------> \(p^2(1-p)\)


Since this is an "OR" case, add the three scenarios together, and do the algebra.

\([p^2] + [p^2(1-p)] + [p^2(1-p)] < p\)

\(p^2 + 2p^2(1-p) < p\)
\(p^2(1 + 2(1-p)) < p\)
\(p^2(3-2p) < p\)
\(p(3-2p) < 1\)
\(2p^2 - 3p + 1 < 0\)
\((p - \frac{1}{2})(p - 1) < 0\)

∴ \(\frac{1}{2} < p < 1\)


And then you know the given information is sufficient if it tells you whether "p" is between \frac{1}{2} and 1
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Mike recently won a contest in which he will have the opport 22 Sep 2018, 01:11
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I looked at it as follows:

we're comparing prob of success = p under two scenarios : when mike takes one shot and when mike takes 3 shots

prob of success = p
prob of failure = 1 - p = q

when comparing one attempt with three : value of p in the single shot vs at least one p in the three shots ~ (1 - q^3)

Stmt 1 gives mixed results when we try different values for p in the above sentence ~ p vs (1 - q^3)
Stmt 2 gives us clear results when we try at least 0.6 in p vs (1 - q^3)

Hence, B

criticism is appreciated
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Mike recently won a contest in which he will have the opport 31 Jul 2019, 21:10
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ctrlaltdel
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
Show more

I think my approach is completely out of the box and for those who have not brushed themselves yet with probability problems.

I used logic and solved this in less than a minute.

Question asks Would Mike have a better chance of winning if he chose to attempt 3 free throws? Clearly if we can get a definite Yes or No, we get our answer.

Statement 1
p < 0.7

P could be anything from 0 to 0.699
If its 0, probably to attempt 3 free throws would be a better chance at winning; BUT if its 0.699 then we really don't know - attempting 1 throw or 3 free throws (out of which 2 should hit) would be a better shot at winning?

Statement 2:
p > 0.6
Definitely p has to be minimum 0.61 and maximum 1 : clearly attempting one throw would be a better chance to win because with respect to the occasional throw and miss, mike's probability of hitting the target by attempting one throw is much higher than when he attempts 3 free throws.
Sufficient
B is the answer.


I request Moderators to review my explanation and suggest me if using logic in such questions is ALWAYS useful or a risky approach in the actual GMAT exam?
chetan2u, VeritasKarishma, Bunuel, gmatbusters, amanvermagmat
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Mike recently won a contest in which he will have the opport 25 Nov 2019, 13:55
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Bunuel
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.


Dear Banuel

I interpreted this - "Mike occasionally makes shots and occasionally misses shots." as the probability of hitting and missing being 50-50. Is that wrong? I got the ans nevertheless as .6 is greater than the overall probability of hitting in exactly 2 shots.

THANKS IN ADVANCE

Answer; B.
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Mike recently won a contest in which he will have the opport 27 Mar 2020, 07:52
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Bunuel I solved it in 3:30 with the same method, how do I solve it in 2:30?
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Mike recently won a contest in which he will have the opport 11 Mar 2025, 06:17
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Question states that Mike occasionally makes shots and occasionally misses them. Thus p ≠ 0 or 1.
Monisha2402
What if the probability of making a successful free throw is 1, in that case option B, p>0.6 cannot be the answer, since the question is really asking if 0.5<p<1 ?
Please correct me if my understanding is wrong..
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Mike recently won a contest in which he will have the opport 11 Mar 2025, 06:26
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grimm1111
aakash1618

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.

Another way to arrive at this is just to work out the cases. If he has 3 shots and must make two, then he can win if:

HH (He makes the first two shots. Third shot is not needed, he already won.)
NHH (Missed the first, made the second two)
HNH (Made the first and the last)

the probabilities of these three winning scenarios in terms of "p" are:

HH ---> \(p*p\) --------------------> \(p^2\)
NHH ---> \((1-p)*p*p\)--------------> \(p^2(1-p)\)
HNH----> \((1-p)*p*p\)--------------> \(p^2(1-p)\)


Since this is an "OR" case, add the three scenarios together, and do the algebra.

\([p^2] + [p^2(1-p)] + [p^2(1-p)] < p\)

\(p^2 + 2p^2(1-p) < p\)
\(p^2(1 + 2(1-p)) < p\)
\(p^2(3-2p) < p\)
\(p(3-2p) < 1\)
\(2p^2 - 3p + 1 < 0\)
\((p - \frac{1}{2})(p - 1) < 0\)

∴ \(\frac{1}{2} < p < 1\)


And then you know the given information is sufficient if it tells you whether "p" is between \frac{1}{2} and 1
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There are some calculation errors:
\(2p^2 - 3p + 1 < 0\)
\((p - \frac{1}{2})(p - 1) < 0\)

The inequality sign at the start of the equation is wrong, it should be greater than ( > ). Otherwise the last few steps are not correct.
\(p(3-2p) < 1\)
After the above step, the solution would be:
\(3p - 2p^2 < 1\)
\(0 < 1 - 3p + 2p^2\)
\(p^2 - 1.5p + 0.5 > 0\)
\(p^2 - 1p - 0.5p + 0.5 > 0\)
\(p(p - 1) - 0.5(p - 1) > 0\)
\((p - 1)(p - 0.5) > 0\)

This is true when p > 1 or p < 0.5.
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Mike recently won a contest in which he will have the opport 04 May 2025, 00:55
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hello, How is (p-1/2)(p-1)<0 translating to 1/2<p<1?
Bunuel
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.
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Mike recently won a contest in which he will have the opport 04 May 2025, 01:41
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NoeticImbecile
hello, How is (p-1/2)(p-1)<0 translating to 1/2<p<1?
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This is an important concept of inequalities that you must get acquainted with for easier solving of problems.

Please go through this compiled concepts on inequalities for better understanding:
https://gmatclub.com/forum/inequalities ... 06653.html

Now, on to the query in hand:

\((p - 0.5)*(p - 1) < 0\)

LHS has 2 terms whose multiplication is less than 0.
This can only be possible if one of them is positive and the other one is negative.
If both terms are positive, then their product should also be positive, then LHS > 0, but this not possible.
If both terms are negative, then -ve * -ve = +ve, then again LHS > 0, but this is not possible.

Case 1: 1st term is negative, 2nd term is positive
\((p - 0.5) < 0\) and \((p - 1) > 1\)
\(= p < 0.5\) and \(p > 1\)
This situation is not possible as p cannot be less than 0.5 and greater than 1 at the same time.

Case 2: 1st term is positive, 2nd term is negative
\((p - 0.5) > 0 \) and \((p - 1) < 0\)
\(= p > 0.5\) and \( p < 1 \)
= \(0.5 < p < 1 \)

If you analyse it a bit, you will find that LHS has roots as 0.5 and 1, i.e. when p = 0.5 or 1, LHS becomes 0. These 2 values of p are important points around which the value of LHS changes.

If you put it on a number line, all values of p before 0.5 make the whole term of LHS > 0, all values of p after 1 make the whole term of LHS > 0.
At 0.5 and 1, LHS = 0.
And between 0.5 and 1, LHS < 0 - this is the case we need as per the problem.
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