Data Sufficiency

MBAsa wrote:Guys, how do you solve this in under 3 minutes

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

Dear MBAsa,
I'm happy to respond to this.

First, let me say, if the difficulty of questions possible on the GMAT goes, say, from 1 to 100, this question is at a difficulty of 2417. You could have a thousand people take the GMAT every day for year, and you would never have any of them running across the advanced concepts in this question.

Let's say that Mike's probability of making one free throw, on one trial, is p. Then, obviously, in the one-shot option, his probability of success is P = p. Easy.

The probability of making at least two of three. This calculation gets us into something called the binomial probability distribution. This would be a meat & potatoes topic on, say, the AP Statistics exam, but the GMAT never tests this.
P(X >= 2) = P(X = 2) + P(X = 3)
The probability P(X = 3), making all three free throws, is easy. That just
P(X = 3) = p^3
The P(X = 2) probability gets us into the binomial stuff. This is two successes and one failure --- p*p*(1 - p), but we have to keep in mind, the pattern of two successes and one failure can happen in 3C2 = 3 different ways ----- SSF, or SFS, or FSS. Thus:
P(X = 2) = 3(p^2)(1-p)
and
P(X >= 2) = P(X = 2) + P(X = 3)

P(X >= 2) = (p^3) + 3(p^2)(1-p)

That just gets us the formulas. Now, the really hard part: we want to compare this cubic function to the function p, which is a straight line. Holy smokes! I can do a ton of math in my head, but it's really beyond me how to do this without using computer graphing capabilities!! The curve below shows the graphs --- it was not intuitively obvious to me that they should intersect at (0.5, 0.5), although now that I see this, I can numerically verify it. The straight purple line is the probability of winning on the one-shot plan, and the green curve is the probability of winning on the get 2-of-3 plan.

Now that we can look at the graph, clearly if p < 0.7, we know nothing --- either curve could be higher. BUT, if we know p > 0.6, we know the green curve is always higher in this region. With the graph, we can see [spoiler](B)[/spoiler] is the answer.

My advice: yes, it's fun to play around with these harder problems, and if they really intrigue you, take a college level stats course some time --- it could only help you in the business world! BUT, for the GMAT, don't worry about solving something like this, with no calculator or graphing capabilities, in under 3 minutes. Unless you can make genetic modifications to yourself, so that you gain the abilities of either Issac Newton or Will Hunting, then I wouldn't worry about doing this without computer graphic capabilities.

Does all this make sense?
Mike