By: Rich Zwelling, Apex GMAT Instructor

Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

*Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?*

*A) 1/24
B)*

*1/8*

*C) 1/4*

*D) 1/3*

*E) 3/8*

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics).

**Explanation #1:**

First, we’ll tackle **pure probability**. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with *incorrect* envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_**A**_ _**B**_ _**C**_ _**D**_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_**1/4**__ _**B**_ _**C**_ _**D**_

We now desire letter B to be placed in an *incorrect* envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_**1/4**__ _**2/3**_ _**C**_ _**D**_

We then desire letter C to also be placed in an *incorrect* envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_**1/4**__ _**2/3**_ _**1/2**_ _**D**_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_**1/4**__ _**2/3**_ _**1/2**_ _**1**_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x ~~2~~ x 1

———–

4 x 3 x ~~2~~

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

**Explanation #2:**

So what about a **combinatorics approach**?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

**_4_ _3_ _2_ _1_ = 4! = 24 possible outcomes**

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes).

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left:

___ ___ ___

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a *mismatch*:

_2_ ___ ___

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_ _1_ ___

And finally, there would be only one choice left for the final envelope:

_2_ _1_ _1_

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed *incorrectly*:

_2_ x _1_ x _1_ = 2 possible outcomes.

But as before, we must consider that *any* of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is **8/24 = 1/3**.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

*If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?*

*A) 1/6
*

*B) 1/3*

*C) 1/2*

*D) 2/3*

*E) 5/6*

Permutations and Combinations Intro

A Continuation of Permutation Math

An Intro To Combination Math

Permutations With Repeat Elements

Permutations With Restrictions

Combinations with Restrictions

Independent vs Dependent Probability

GMAT Probability Math – The Undesired Approach

GMAT Probability Meets Combinatorics: One Problem, Two Approaches