Some of the GMAT’s hardest Problem Solving problems can be made exponentially easier by keeping a famous JayZ lyric in the back of your mind. When you hear the phrase:
If you’re having girl problems, I feel bad for you son?
What immediately springs to mind?
I got 99 problems but a b**** ain’t one.
Now, what’s the GMAT genius in Hova’s lyric? He didn’t tell you what his problems WERE, he just told you what they WEREN’T. Explaining 99 problems would take way more than the two minutes you’d have for a quant problem or the ~3 minutes that Jay wants to spend on a track. And, like JayZ, you want to be Mr. One Take on GMAT problems, doing things the efficient way and getting to the answer much more quickly. So heed his advice when you see a problem like:
Solange takes four roundhouse swings at her brotherinlaw. If she is just as likely to connect on any one punch as she is to not connect on that punch, what is the probability that she connects on at least one punch?
Now, there are plenty of sequences in which she can connect:
Hit, Miss, Hit, Miss
Miss, Miss, Miss, Hit
Hit, Hit, Hit, Hit (ouch!)
etc.
Trying to list out all the different ways in which she can land a punch is almost as timeconsuming as listing all of one’s 99 problems. But think of it this way – which of the sequences available “ain’t one”; which ways does she NOT land a punch. There’s only one:
Miss, Miss, Miss, Miss
And so if we’re calculating the probability among the 16 total sequences (each of two things can happen at each of four points, so the total number of sequences is 2^4 = 16), then if one doesn’t work the other 15 must work. So the probability is 15/16. And the “formula” to use on this essentially derives straight from JayZ’s lyrics about what “ain’t one”:
For complementary events (when the probability of A + the probability of B = 100%), the probability of A = (1 – “not A”). And most strategically, this can be used as:
The probability of “At least one” = (1 – probability of “none”)
So if you’re calculating the probability of an outcome that has many different paths, see if it’s a cleaner calculation to determine the number of paths that “ain’t one” of your desired outcomes, and then just subtract those from one.
Note that this ideology doesn’t just extend to probability. In many problems, calculating all the outcomes that “are” desired is a whole lot harder than calculating the outcomes that “ain’t one” of the desired. Consider this problem from this week’s GMATT Mondays session:
Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
A) 66
B) 67
C) 68
D) 69
E) 70
Here look at the answer choices – they’re all very, very high numbers for the range (170) in question. So if your goal is to try to come up with all the possible coin combinations that work, you’ll be there a while. But what about the combinations that “ain’t one” of the possibilities? Since the maximum is 70, if you find the combinations that don’t work you’re doing this much more efficiently…and the answer choices tell you that at maximum only four won’t work so your job just became a lot easier.
With 2 and 5 cent coins as your options, you can’t get to 1 and you can’t get to 3, so those are two “ain’t one” possibilities. And then “100% minus… comes back into play” – Notice too that 70¢ is the maximum possible sum (that would use all the coins), so 70¢ – 1¢, or 69¢, and 70¢ – 3¢, or 67¢ are impossible too. So the answer is 66, but the takeaway is bigger: when calculating all the possibilities looks to be far too timeconsuming, you often have the opportunity to calculate the possibilities that “ain’t one.” You’ve got a lot of problems to tackle on test day; hopefully this strategy allows you to make one question much less of one.
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By Brian Galvin
