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

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

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

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

Bunuel I solved it in 3:30 with the same method, how do I solve it in 2:30?

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