Incredibly Helpful GMAT Tutor NYC & Online

# 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? GMAT Explanation with strategies and more examples!

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

Good old Tanya's letters! This is from the GMAT Prep Tests 1 and 2 so if you haven't tackled those then hold off on working through this.

## Tanya and her letters

This is a tricky GMAT combinatorics probability question that has a somewhat simple solution. And that's something that you need to keep in mind on the GMAT. Not everything is about elegance. Or math power. Or some trick or strategy. A lot of your success on the GMAT will rely on making step by step, organized, practical choices. Not all combinatorics questions need to be tackled with factorials or the slot method.

Some questions are best solved by writing out scenarios and/or spotting a pattern. Tanya and her letters is one such question. It can be solved purely with combinatorics formulas but most people will probably be more successful at least doing part of this one with a more practical approach.

So what's a sign for: don't cram it into a formula and stay practical! The biggest sign, there aren't that many possibilities. So writing it out wouldn't be a big time suck (there are only 4 envelopes and 4 letters). Secondarily, if the setup feels non-standard or has a bunch of moving parts that don't fit well into a formula that might indicate that a more practical/non-math setup would make more sense.

## Let's start by defining the question

what is the probability that only 1 letter will be put into the envelope with its correct address?

OK - no problem. Probability doesn't mean difficult. We need, Our Scenario/Total. So, only 1 letter will be put into the envelope with its correct address/Total possibilities.

## Now go ahead and start setting things up/organizing information

Letters: A, B, C, D

Addresses/Envelopes: a, b, c, d (I used lowercase versions of the letters to make it easy to match up correct letters/addresses)

The total is usually the easiest thing to figure out because you likely won't have any constraints that could make the setup trickier. In this case we have 4 options for letters and 4 options for addresses. So you have 4 spaces for letters (treat this as an ordering question). How many different ways can you arrange 4 things? 4! = 24.

Now we need to calculate the numerator which is the specific scenario: only 1 letter will be put into the envelope with its correct address

For this you might just write it out. It's quick. You should see the pattern after the first two so you don't have to write out the rest.

Aa Bd Cb Dc

Aa Bc Cd Db

Bb Ac Cd Da

Cc Ab Bd Db

Dd Ac Ba Cb

Dd Ab Bc Ca

## You can also do this with a formula which is neater but not necessarily faster or easier/less prone to error.

There are 4 correct matchups:

Aa

Bb

Cc

Dd

And then you need to figure out the remaining incorrect ones. So, let's just take the first correct one Aa. For the next slot, b you have 2 options (c, d) but once you've chosen c or d you don't have choices for the last two slots, you're locked in, you only 1 option each because any other arrangement will create a "correct" address/letter combination.

So you have 4 correct options (Aa, Bb, Cc, Dd) and then 2 options for the incorrect ones. 4*2 = 8.

That's not complicated but I see a lot of GMAT tutoring students get tangled up there. And considering you don't have to write out much to see the pattern, a practical approach is usually more successful on this question.

## Finally, let's put it all together

8/24 = 1/3.

So 1/3 of the time you are left with only one of Tanya's letters in the envelope with the correct address.

## More GMAT Combinatorics and Probability Questions!

Here's a great one from the GMAT official guide that also touches on the idea of being practical with combinatorics: A researcher plans to identify each participant in a certain medical experiment

And another tough oddball combinatorics question from the GMAT Prep Tests: A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors

And here are two tough official GMAT probability questions:

A basket contains 5 apples, of which 1 is spoiled and the rest are good.

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of , where b is an integer?

# Among a group of 2,500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from the 2,500 people, what is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks? GMAT Explanation

Among a group of 2,500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from the 2,500 people, what is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks?

(A) 9/50
(B) 7/25
(C) 7/20
(D) 21/50
(E) 27/50

Combinatorics questions, especially ones tangled up with probability, scare students. Can there be a very challenging probability question? Yes! Are most of them tough? Not really. Do you need a bunch of complex math to solve GMAT probability questions? Nope. Let's see what we need to get this one solved.

## Define the question

What is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks?

OK, not so bad. Munis but no oil stocks. So what is probability exactly? Specific Scenario/Total Scenarios. So Muni/Not Oil/Total. This is starting to feel like overlapping sets with a side of probability...

## Let's get the information organized!

Let's call the 2500 people 100 people. It won't make a difference in the answer because everything is based on percents/proportions and will make it easier to do the calculations. So:

Municipal Bonds: 35

Oil Stocks: 18

Muni and Oil: 7

Looks like we have a two group overlapping set. We're going to need a diagram.

So just fill in the given information and then make easy inferences. You'll be able to get the whole chart filled in. You end up with 28 people that have municipal bonds but not oil stocks. But, 28 isn't the answer because we're asked about probability. So take 28 and put it over the total. What's the total? The 2500. BUT we changed that to 100. 28/100 = 7/25.

## More GMAT two group overlapping sets and probability practice!

Here's a challenging sets question from the GMAT Prep Tests 1 and 2. Stay organized. Make the easy inferences and you should be fine.

In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:

Here's one from GMAT Question of the Day that's very similar in that the answer is based on proportions. It doesn't have a probability component but again: same idea and great practice.

GMAT Question of the Day 2 group overlapping sets proportions and percents

This one is a bit different but will still give you 2 group practice and will give you exposure to overlapping sets on data sufficiency which often comes down to counting equations.

GMAT Question of the Day 2 group overlapping sets Data Sufficiency

And here's a very tough DS overlapping sets questions. Again, deals with counting equations/system of questions.

GMAT Question of the Day Challenging 2 Group Overlapping Sets Data Sufficiency Counting Equations

A basket contains 5 apples, of which 1 is spoiled and the rest are good. If Henry is to select 2 apples from the basket simultaneously and at random, what is the possibility that the 2 apples selected will include the spoiled apple?

(A) 1/5
(B) 3/10
(C) 2/5
(D) 1/2
(E) 3/5

GMAT tutoring students tend to dislike probability questions. A couple of things to remember regarding GMAT probability:

1. It doesn't come up that much.
2. When it does come up often the question isn't that challenging.
3. If you do get a challenging probability question you can always skip.

Let's focus on the question first: what is the possibility that the 2 apples selected will include the spoiled apple?

We need a spoiled apple selected. OK. There's a very practical way of doing this:

G1 G2 G3 G4 S

10 Possible pairs. 4 of them have a spoiled apple. So 4/10 or 2/5 chance of having a pair with the spoiled apple.

G1 G2

G1 G3

G1 G4

G1 S

G2 G3

G2 G4

G S

G3 G4

G3 S

G4 S

Easy. If the numbers are small there's nothing wrong with writing things out. In fact, it can be the best approach.

The other way to do it is with the slot method. Calculate the number of ways to create a group of two from 5 things. That will be your total or your denominator. And then calculate the number ways to create a group of two with the constraint that one of those things (the spoiled apple) must be included. That will be your numerator. Remember that probability is just specific scenario/all scenarios.

5*4/1*2 = 10 (the number of ways to create a pair from 5 things)

With one of the spots reserved for the spoiled apple you're only left with the other spot to populate. So how many things can go in that spot? 4 (the 4 good apples). So the numerator is 4 and the denominator is 10. 4/10 = 2/5.

## More Challenging GMAT Combinatorics/Probability Examples

Here's an even more challenging GMAT Combinatorics/Probability question with an in depth explanation:

Tanya prepared 4 different letters to be sent to 4 different addresses.

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of , where b is an integer?

(A) 1/2

(B) 1/3

(C) 1/4

(D) 1/5

(E) 1/6

Be super comfortable with difference of squares. You're almost guaranteed to need it on your GMAT. You need to know it forwards, backwards, and upside-down. There isn't much content you need to know for the GMAT quant but in order to be consistently successful so that you can go in there on test day and snag your 700+ score you need complete fluency with that limited content. On this one we're looking for a product of two terms that will produce difference of squares. So we need something in the format: (x+y(x-y). There's only one pairing that gives us the right format. Now you have to figure out what that means in terms of probability. In general, probability is specific scenario/total scenarios. You can calculate the total and the specific scenarios using the slot method. How many teams of 2 can you make from four things? (4*3)/(1*2) = 6. So that's your denominator. And then there's only one team that works, the pairing of (x + y)(x - y). So that leaves you with 1/6. E. Detailed diagrams below and an in depth video explantation here: If two of the four expressions

You also can think about this in terms of straight probability and say: 2 choices our of 4 and then 1 choice out of 3. 2/4 * 1/3 = 2/12 = 1/6. Comment with any questions or additions. Happy studies!

## More GMAT probability, combinatorics examples and explanations

Here's a tough one from the GMAT Prep tests. It's even a bit more challenging than the one above but has a very practical solution that should help focus your GMAT thinking: Tanya prepared 4 different letters to be sent to 4 different addresses.

# GMAT Question of the Day - Data Sufficiency - Probability

Roy has a bowl with 12 red and j blue marbles. If Roy picks two random marbles out of the bowl it is more likely that he will pick one of each color than only two blue marbles. Are there an equal number of red and blue marbles in the bowl?

(1) j > 11

(2) The chance of picking two blue marbles is greater than 5/23

## GMAT Question of the Day Solution

On GMAT Data Sufficiency questions take your time to understand the given information. Avoid rushing into the statements. In this case you need to unravel this statement: If Roy picks two random marbles out of the bowl it is more likely that he will pick one of each color than only two blue marbles. This is critical. Now you might be thinking that this question is very open ended because you don't know how many blue marbles you have but the given information severely limits the options. The number of blue marbles must be less than or equal to 25. Here's the math to calculate that.

This is based on this statement: If Roy picks two random marbles out of the bowl it is more likely that he will pick one of each color than only two blue marbles.

j/(12+j) * (j-1)/(12+j-1) *(2!/2!)  < j/(12+j)*12/(12+j-1)*(2!/(1!*1!))

j<25

Statement (1) Could be a Yes, 12, or a no with anything greater than 12. Statement (2) give you the same information but in a slightly different way. We need to know whether we have 12 blue marbles so just try plugging in 11 marbles or 12 marbles see which one (or maybe both) agrees with the information in the statement. We can see that when picking 11 you get exactly 5/23. So there must be more than 11 blue marbles. That is the same info as statement 1. You could have the same number of marbles 12 and 12 or you could have more blue marbles, up to 24.

# CONTACT

Atlantic GMAT Tutoring

405 East 51st St.

NY, NY 10022

(347) 669-3545