This is the third in a series of probability articles for the GMAT Quantitative Section. In the first, I discussed the “AND” and “OR” rules for probability. In the second, I discussed the “complement rule” and how to use this to solve “at least” problems in probability. For a warm-up, here are some challenging GMAT Problem Solving problems on probability.

1) Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?

- 1/30
- 1/15
- 1/5
- 2/5
- 7/20

2) A division of a company consists of seven men and five women. If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?

- 1/6
- 2/5
- 2/9
- 5/12
- 5/33

3) John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?

- 1/2
- 2/5
- 3/10
- 7/20
- 8/45

Do these problems make your head spin? Then you have found just the post you need!

## Probability and counting

Fundamentally, the definition of probability is

The previous two posts talked about various tricks for calculating probabilities in different scenarios, but in some problems, we just have to count the numbers in the numerator and the denominator. This means, we need to understand basic counting techniques, including combinations and permutations. The details of the counting techniques are explained in these posts. Any probability problem involving counting is really two counting problems in one —- we have to calculate the denominator, the number of all possible cases, and the numerator, the number of only those cases that meet the condition specified. I’ll just mention that often, the most challenging part of any counting problem, especially for special cases (e.g. the numerators of these probability expressions) is how we ** frame** the problem. Details of this process will be explained below in the solutions to problems.

## Summary

If counting techniques are unfamiliar to you, read those two other posts. Once you feel confident with counting techniques, give the problems above another try before reading the solutions below. In the last article in this series, I will discuss a special case of probability problems on the GMAT: geometric probability.

## Practice problem explanations

1) First, we will count all the possible arrangements of the five children on the five seats, all the possible orders. This is 5! = 120. That’s the denominator.

Now, the more challenging part: we have to figure out how many arrangements there are involving C & E sitting together. This is a tricky problem to frame, so I’ll demonstrate the steps to follow. First, let’s look at the seats these two could be next to each other. There are four possible pairs of seats in which they could be next to each other

i. X X _ _ _

ii. _ X X _ _

iii. _ _ X X _

iv. _ _ _ X X

In each of those four cases, we could have either CE or EC, either order, so that’s 4 x 2 = 8 ways we could have just C & E sitting next to each other with the remaining three seats empty.

For the final step, we need to consider the other three children, A & B & D. In each of the eight cases, there are three blank seats waiting for those three, and those three could be put in any order in those blank seats. Three elements in any order —- that’s 3! = 6. Thus, the total number of arrangements in which C & E would be next to each other would be 8 x 6 = 48. This is our numerator.

The probability would be this number, 48, over the total number of arrangements of the children, 120.

Answer = **D**

2) First, the denominator. We have twelve different people, and we want a combination of two selected from these twelve. We will use the formula:

which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers. Here

That’s the total number of pairs we could pick from the twelve employees. That’s our denominator.

For the numerator, we want every combination of two from the five female employees. That’s

That’s the number of pairs of female employees we could pick from the five. That’s our numerator.

Answer = **E**

3) For the denominator, we are going to pick two books from among ten total: a combination of two from ten. Again, we will use the formula:

which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers. Here

That’s the total number of pairs of books we could pick from the ten on the shelf. That’s our denominator.

Now, the numerator. We want one novel and one reference work. Well, there are four novels and two reference works, so by the FCP, the number of ways we can pick this is 4 x 2 = 8. That’s the total possible number of pairs involving exactly one of these four novels and exactly one of these two reference works. That’s our numerator.

Answer = **E**

Greetings Mike,

For question 3, why isn’t 4/10 * 2/10 sufficient enough? What is it that I am missing exactly from the way I was trying to solve the problem?

*also, for question 1, is there a mathematical approach to finding out how many ways the seating can occur? or the best way is to draw it out?

Hope to hear from you soon

Dear Herpal,

I’m happy to respond.

For question 3, these books are being selected “

without replacement” — that is, the first choice is made from a group of 10, but the second choice isNOTmade from a group of 10, because the one that was picked first is missing from the group. Your approach would work if we picked one book, put it back, and then picked another book from the set of 10. That’s not what is happening in this situation.For question #1, I

showa mathematical solution in the text explanation. That sort of logical reasoning IS what mathematics is about. Thatismath! If you mean to ask: is there a simple formula way to plug in? No. Mathematics is not simply about plugging into the right formula. Don’t look for a formula and don’t draw it out. The very best way to solve that problem is the one that I show in the text explanation.Does all this make sense?

Mike

Thanks Mike! It does!

Dear Herpal,

You are quite welcome! Best of luck!

Mike

Hi Mike,

But in question 3, if you account for it being “without replacement”, 4/10 * 2/9, you still get a different answer. What am I missing?Thanks in advance

Dear Felipe,

That’s a great question, and it gets into some subtleties. The counting method I showed is independent of order of selection: it just deals with “finished product” choices of the two books selected. By contrast, when you start thinking about the P of the first choice and the P of the second choice, you are specifying an order of selection. For example, in your calculation, 4/10 * 2/9, that’s the probability for picking one of the novels FIRST and THEN picking a reference work in your SECOND choice from the remaining nine books. The problem is: we want 1 novel and 1 reference work, but we don’t care which one was picked first. This particular probability (novel first, reference work second) is only half the calculation. The full calculation would include selecting the two books in either order:

P = P(novel 1st AND reference 2nd OR reference 1st AND novel 2nd)

= (4/10)(2/9) + (2/10)(4/9)

= 8/90 + 8/90

= 4/45 + 4/45 =

8/45That’s exactly what we got using the counting method. The “without replacement” perspective already implies an order of selection, and if one goes down that route, one has to consider very carefully: are we really concerned with the actual order of selection, or do we care only about the finish result, irrespective of selection order? The counting solution avoids this problem entirely by focusing exclusively on the “finished product,” once selection is already done.

Does all this make sense?

Mike

Yes, i get it now. Thank you Mike!

Dear Felipe,

You are quite welcome, my friend! Best of luck to you in all your studies!

Mike

Great article as always, Mike!

Dear Sheriari,

Thank you for your kind words. Best of luck to you!

Mike

Hi Mike ,

How about solving question 1 as :

Assuming the 2 kids who are sitting together as 1 unit.So now finding out in how many ways this one unit sits taking 2 seats together which is equal to 4/5 ways.

Now that 2 these 2 kids can sit in 2 ways in those 2 occupied chairs,the ways will be 1/2.

Therefore the probability of 2 kids sitting together is 4/5 * 1/2 = 2/5 ways.

Please comment on this approach.

Dear Megha,

With all due respect, I don’t believe this is a reliable method for solving problems. Among other things, you assume the two kids sit together and can be treated as single unit — in doing so, you assume the very conditions whose probability we are trying to determine, so there’s no consideration of all the possibilities in which those two kids aren’t sitting together.

Does all this make sense?

Mike

Hey Mike,

what if I approach prob 1 in a similar but a slightly different approach?

Total # of possible outcomes = 5!

Now we consider the two kids sitting together as 1 unit, then then total number of ways we can arrange all kids would be 4! x 2 (times 2 to account for the order)

so now the probability = (4! x 2)/5! = 2/5. Is this approach correct?

Thanks,

Kriti

Dear Kriti,

Yes, that approach is perfectly correct!

Mike

Dear Mike ,

I tried to solve problem 3 : John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?

P ( i novel & 1 reference ) =

(4/10)*(2/9) = 4/45 if novel is selected before reference book. then I stopped and thought this is the answer ( of course I didnt find it in the answer choice ) . My question is why do we need to do

(2/10)*(4/9)= 4/45 if novel is selected after reference book

then add them 4/45+4/45=8/45

Sally,

Many probability problems can be solved in more than one way. You found another, 100% acceptable solution, to this problem. That’s great! It’s actually very helpful for your mathematical understanding to explore as many different ways to solve a problem as possible — the more possible solutions you have seen, the more tricks you have up your sleeve when you are looking at a new problem.

Does all this make sense?

Mike

Yes it does make sense. Thank you Mike so much

Dear Sally,

You are quite welcome. Best of luck to you!

Mike

Mike for the last two. Can you also solve them this way:

for picking women: out of 12 we have 5 choices and out of 11 remaining we have 4 choices. so

5/12 * 4/11 which simplifies to 5/3 * 1/11 or 5/33

For picking books.

We can pick a novel and a reference book (4/10 * 2/9) or(+) a reference book and a novel (2/10 * 4/9)

Dennis,

Yes, both of those approaches are fine. Remember to add the two pieces in the latter problem — OR means add. Best of luck to you.

Mike

for problem 3 , Why do we need to add the 2 pieces together and why only one is not enough to solve the problem ( I mean pick a novel and a reference book ((4/10 * 2/9) why is it not enough to solve the problem ) ?

Sally,

What you are asking is an excellent question. It’s getting into something extremely subtle about probability. We can solve this question two ways, via “counting techniques”, which I show above, and via “selection process”, which you did. More so than in any other area of math, use of any formula in probability involves a worldview, and different approaches involve different assumptions and perspectives. All this is hard to put into words. Here’s one way to talk about it.

When I used the counting techniques, I was counting physical objects. In that view, counting Novel B and Reference K is the same as counting Reference K and Novel B — either way, I would have the same two physical books in my hands. In some sense, that’s why I think that approach is more intuitive.

When you used the selection process technique, technically, you were routes to get to a final product. Picking Novel B first and Reference K second gives us the same final product as Reference K first and Novel B second, but it’s a different route, and the routes are what we are counting. In general, calculations involving the “and” and “or” rules, to some extent, involve counting “routes” one can take to the desired final outcome. I can pick (B, then K) OR (K, then B), and that “OR” is a real mathematical requirement completely inherent to that approach to the problem.

I realize this is probably not a completely satisfying answer. To some extent, you need to solve probability questions (ideally, each one in multiple ways), read solution for them, and develop you own intuition over time. If you try to understand probability completely via left-brain rule-based logic, you will be disappointed. There’s an irreducible right-brain element, and this is the hardest thing to specify or communicate in words.

Does all this make sense?

Mike

Hi Mike !

I solved 2nd and 3rd question in a different way. Please comment on the method:

John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?

(4/10)*(2/9) = 4/45 if novel is selected before reference book.

(2/10)*(4/9)= 4/45 if novel is selected after reference book

so 4/45+4/45=8/45

Dear Hashir,

That’s also a perfectly fine way to solve this. Many GMAT Quant problems have more than one solution, and recognizing the different solutions to an individual problem is a clear path to developing mathematical mastery. See:

http://magoosh.com/gmat/2013/multiple-solutions-in-gmat-math/

Mike

Thank You Mike. Much appreciated.

Hashir,

You are quite welcome. Best of luck to you.

Mike

good stuff.

Dear Maggie,

Thank you for your kind words. Best of luck to you!

Mike

I’m assuming problem (1) – D is 2/5, not 2/15.

KC –

Yes, thank you for pointing out this typo. I just corrected it.

Mike