Tackling some of the tougher GMAT probability questions efficiently relies on both steady practice and your ability to make two key decisions well. First, you will need to quickly and accurately assess the total number of possible outcomes (the denominator of your probability equation). Second, within a multitude of possible approaches, you will need to determine the most efficient route to calculate the number of desired outcomes (the numerator of your probability equation).

With the clock ticking away on your GMAT CAT, figuring out the total number of possibilities can be time-consuming and fraught with room for error. For instance, if a question asks about the probability of getting at least 2 heads on 5 coin tosses, you could sit there all day writing out possibilities:

HHTTT

HTHTT

HTTHT

So forth and so on. I know I got dizzy with the possibilities just writing those three out. There is a better and more efficient way. For every coin that you toss there are 2 possibilities. You can think of the total possibilities like a permutation problem.

__2__ __2__ __2__ __2__ __2_

1^{st} 2^{nd} 3^{rd} 4^{th} 5^{th}

Just like in a GMAT permutations question when we are trying to determine the total number of codes possible or 4-digit numbers, we would multiply these individual probabilities together. Therefore, there are 2x2x2x2x2 = 2^5 = 32 total possibilities.

Next, we need to look at the numerator (desired outcomes). We want to find the all of the possibilities that have at least 2 heads, which means that we could have 2 heads, 3 heads, 4 heads, or 5 heads. To do so, we would need to count all of the different ways that these possibilities could be arranged. Again, we find ourselves in a situation that will be time-consuming and fraught with error. Instead of going down this path, remember that the sum of the probabilities of a complete set of mutually exclusive possible outcomes is 1. Thus, as is often the case on “at least” probability questions, we can look for those options that are restricted. Then we only have to count the options that have 1 or 0 heads.

TTTTT

HTTTT

THTTT

TTHTT

TTTHT

TTTTH

There are only 6 of those, instead of the 26 possibilities the other way.

Finally, we can either subtract 6/32 from 1 in order to remove all of the restricted possibilities from 1 or we can subtract 6 from 32 and use the result as the desired possibilities. Either way, the answer is 26/32, which you can reduce down to 13/16.

Let’s look at another to make sure we have this down.

*Question*

A test has 5 multiple-choice questions. Each question has 4 answer options (A,B,C,D). What is the probability that a student will choose “B” for at least four questions if she leaves no questions blank? Pause a moment and try it for yourself fir*st.*

*Step 1: Total number of possibilities*

There are 5 questions and each has 4 possibilities, so our total possibilities would be 4x4x4x4x4 = 4^5 = 1024

*Step 2: Approach Desired Strategically*

Here there are far more possibilities for 0, 1, 2, or 3 “B’s,” so let’s get a total for 4 or 5 “B’s”.

All B’s — B,B,B,B,B

Four B’s — A,B,B,B,B – B,A,B,B,B – B,B,A,B,B – B,B,B,A,B — B,B,B,B,A

C,B,B,B,B – B,C,B,B,B – B,B,C,B,B – B,B,B,C,B – B,B,B,B,C

D,B,B,B,B – B,D,B,B,B – B,B,D,B,B – B,B,B,D,B – B,B,B,B,D

3 x 5 = 15 because we can repeat the same pattern for each letter other than B

We can also calculate the total possibilities of 4 B’s by calculating the possibilities for each “no-B” position.

**No-B,**B, B, B, B = **3**x1x1x1x1 = 3

B, **No-B**, B, B, B = 1x3x1x1x1 = 3

B, B, **No-B**, B, B = 1x1x3x1x1 = 3

B, B, B, **No-B**, B = 1x1x1x3x1 = 3

B, B, B, B, **No-B** = 1x1x1x1x3 =3

A total of 15 possibilities with 4 B’s in the mix

That gives a total of 16 different ways that a student can choose at least 4 B’s here.

16/1024 = 1/64 as our final probability.

Keep these two decisions in mind each time that you approach a tough probability question on the GMAT quantitive section. You don’t have to write out all of the possible outcomes in order to tackle these on test day!

**The language of probability can take a while to learn, especially if you are unfamiliar or out of practice with it to start. Post your questions below, and we can help you get on track.**

Serina Isch has taught for Kaplan since 2009. As a former high school English teacher, she started with the pre-college tests ACT, SAT, and PSAT but soon transitioned to grad with the addition of the GRE to her repertoire. Since that time, she also has added the GMAT and LSAT and has scored at the 95th percentile or higher in each test that she teaches. She is a full-time Kaplan instructor who teaches and tutors both on-site and in the anywhere classroom. Serina was named the Kaplan’s Oklahoma Teacher of the year in 2009, 2010, and 2011 and was named Kaplan’s Grad South Region Teacher of the Year in 2010. She holds a BA, summa cum laude, in English from Oklahoma State University and is currently finishing an MA in Technology in Education from Teachers College, Columbia University. When she isn’t helping students get into the grad program of their dreams with Kaplan or working on her own degree, Serina loves to hang out with her two kiddos and her husband Chris, volunteer at her kids’ elementary school, and read anything she can get her hands on.

Hi,

Can I ask why in the first example you use permutations and no combinations? I understand that they are not giving it any order, so why not using combination ?

Thanks