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Mike recently won a contest in which he will have the opport

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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 28 Jan 2017, 21:33
anairamitch1804 wrote:
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 ?
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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 29 Jan 2017, 03:47
gmatkiller666 wrote:
anairamitch1804 wrote:
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 ?


(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

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

Hope this helps.
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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 27 Mar 2017, 13:34
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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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 27 Jul 2017, 00:41
aakash1618 wrote:
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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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 22 Sep 2018, 01:11
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  [#permalink]

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New post 31 Jul 2019, 21:10
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 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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Re: Mike recently won a contest in which he will have the opport  [#permalink]

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New post 25 Nov 2019, 13:55
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 = \(C^2_3p^2(1-p)+p^3=3p^3(1-p)+p^3=p^2(3-2p)\).

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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Re: Mike recently won a contest in which he will have the opport   [#permalink] 25 Nov 2019, 13:55

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