GMAT
PROBABILITY
Note: in order to do well in probability, you MUST already have an understanding in
permutations and combinations. Please work through those chapters before this one.
Introduction
Probability is often seen as the real evil of the GMAT. But the truth is that by mastering one simple
fraction, you can make probability very approachable and even easy. As with permutations and
combinations, we will ignore most of the probability formulae you remember from math class and just
focus on what’s really happening. Once you understand that, everything else just falls into place.
Tactical Tip: Even if are skilled in probability questions, the techniques we teach here are highly
efficient and specifically tailored to GMAT questions. Everyone should benefit from these strategies
and techniques.
I. Probability Made Simple
Let’s say you have a single six-sided die. If you role it, what is the probability you will roll a 5?
There are 6 sides to the die, and each one could come up. The 5 side is one of those sides, right? So
there are 6 possible outcomes, and the five is one of them. Therefore, there is a 1 in 6 probability that
the five will come up.
Strategic tip: The Bottom and the Top
If you want to make probability approachable, just think of it as a fraction, and work Bottom to Top.
The bottom number is the total number of possibilities that could happen, while the top number is the
number of possible ways to achieve the desired result.
Or, more simply, P = what we want/all of what’s possible.
Apply that to the above example:
Bottom: 6 outcomes are possible
Top: 1 outcome we want
Probability: 1/6
Let’s change the above example just a bit to see this in action. Now, we still have one die, but we want
to know the probability of rolling a 5 or a 6?
Now, there are still six different possible outcomes, right? Of those, we want a 5 or a 6. Therefore,
two outcomes give us what we want. Think Bottom to Top:
Bottom: 6 outcomes are possible
Top: 2 outcomes we want
Probability: 2/6 = 1/3
Here are several other examples to help you understand the basics of probability:
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck
will produce a 4?
Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 4 cards give us a 4
Probability: 4/52 = 1/13
Example 2
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck
will produce a black card?
Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 26 cards are black
Probability: 26/52 = ½
Example 3
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck
will produce the Ace of Spades?
Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 1 Ace of Spades
Probability: 1/52
II. Multiple Events: Combining Probabilities
Note: this section is oriented to 35 and over scorers on the math section.
Now you understand that probability is just a fraction. But we have to take this further. Many
probability problems involve more than one event happening at once. For questions involving multiple
events, the answer combines the probabilities for each event. This can be confusing, but with a little
extra logic and understanding, you will be able to tackle them.
The only vocabulary you’ll ever need: “And” and “or”
There are two ways events can happen together in the same probability problem: either they could both
happen separately or they must happen together. We will discuss each case separately below.
Scenario 1 – “or”: If both events do not necessarily have to occur together, an “or” may be used as in
“I will be happy today if I in at bingo or receive an email.” These are two separate events, independent
of each other, and the outcome of one event does not influence the outcome of the other event. Most
significantly, the sum of all outcomes does not add up to 1.0000000
In this statement, it is clear that two different things will make you happy. Your chances of getting
email are much better than your chances of winning at bingo, but it doesn’t matter. If either will make
you happy, then you don’t need to win at bingo to make you happy. But if you happen to win at bingo
and not get email, you’ll still be happy. It’s a win-win situation. Your chances of being happy, then,
are higher than your chances of either winning at bingo or getting email are alone. Therefore, if your
chances of winning at bingo are 1:50 and chances of getting an email that day are 3:4 then your
chances of getting an email or winning at bingo are clearly larger than merely the 3 out of 4 chance of
getting the email. But do we merely add them together?? No, we can’t do that – and you should
understand why.
2
Let us give a simple example of this principle:
Suppose we had four separate events, each of which would make us happy, and each carrying odds of
3:4 of happening, then by adding four separate chances of 75% together, we would arrive at a 300%
success rate. That can’t be – as you know, all probabilities are expressed as a fraction between zero and
one.
The way to compute this last problem is to figure out the chances that we would NOT be happy
(meaning that none of the four events making us happy, would occur)
The chances of this happening is the product of ¼ x ¼ x ¼ x ¼ or 1/256
So we have one chance out of 256 possible outcomes that neither of the four events making us happy,
would occur. By corollary, all other outcomes would make us happy, as we would be assured that at
least one, perhaps more, events would occur making us happy.
Consequently, our rate of “success” – namely making us happy, is 1 – 1/256, or 255/256
Scenario 2: “or”
In this case of “or,” we add probabilities together to get a higher overall probability,
provided the sum of the chances of all outcomes add up to 1.
Example 4
John will win $100 if, from a deck of 52 standard playing cards, he chooses either a 7 or a 9 when
pulling a single card from the deck. What is the probability that John will win $100?
Example 4
John will win $100 if, from a deck of 52 standard playing cards, he chooses either a 7 or a 9 when
pulling a single card from the deck. What is the probability that John will win $100?
Answer:
Start by noticing the word “or” in the question. How can John win? He can win by pulling out either a
7 or a 9. His chances of doing that are higher than if he could win only by pulling out a 7. In that case,
he’d only have 4 cards that would make him win $100, now he has 8 cards. To find the total
probability, we need to figure out the probability of each event and then add the together.
So, what is the probability of each? Think Bottom to Top:
Probability of Choosing a 7:
Probability of Choosing a 9:
Bottom: 52 outcomes are possible
Bottom: 52 outcomes are possible
Top: 4 cards are 7’s
Top: 4 cards are 9’s
Probability: 4/52=1/13
Probability: 4/52=1/13
Total probability of choosing a 4 or a 7: 1/13 + 1/13=2/13.
3
Example 5
A fair die is tossed once. What is the probability the die will land on a 2 or a 5?
Answer:
Probability of landing on a 2:
Bottom: 6 outcomes are possible
Top: There is only 1 two.
Probability: 3/6
Probability of landing on a 5:
Bottom: 6 outcomes are possible
Top: There is only 1 five.
Probability: 1/6
Total probability of landing on a 2 or a 5: 1/6 + 1/6 = 2/3 = 1/3
Scenario 2 – “and”: If two events have to occur together, generally an "and" is used. Take a look at
this statement: "I will only be happy today if I get email and win the lottery." The "and" means that
both events are expected to happen together.
Your chances of getting email may be relatively high compared to your chances of winning the lottery,
but if you expect both to happen, your chances of being happy are slim. Unlike in scenario 1, here we
expect both things to happen simultaneously, not one or the other. In that case, we should expect the
probability to reduce.
In the case of “and,” we multiply probabilities together to get a lower overall probability.
Example 6
If a coin is tossed twice, what is the probability that on the first toss the coin lands Heads and on the
second toss the coin lands Tails?
Answer:
First note the "and" in this problem. That means we expect both events to occur together, and that
means fewer options, a less likely occurrence, and a lower probability.
If this coin does what the problem says it will do, the coin toss will look like this:
HT
The probability that the coin will land on Heads, you should now understand, is ½. The probability
that the coin will land on Tails is also ½.
Expect the answer to be less than the individual probabilities of either event A or event B, so less than
½. Since we want them to happen together, we multiply individual probabilities. ½ × ½ = ¼.
Example 7
What is the probability that rolling two identical dice together will result in two fives?
Answer:
Note: not all “and” questions include the word “and”. They may just imply it. In this case, to
get the result of two fives, the first die must be a five and the second one must be a five. We have
the “and” without seeing it.
Probability first roll is 5: 1/6
4
Probability second roll is 5: 1/6
Probability both are 5: 1/6 x 1/6 = 1/36
III. More about “and” problems: Independent and Dependent Events
Whenever two or more events happen at the same time, i.e., when we use the word “and,” you will
have to decide if the events are independent or dependent.
Examples 6 and 7 are examples of Independent probabilities. The outcome of the first event does not
affect the probability of the second. Coin tosses are independent. They cannot affect each other's
probabilities; the probability of each toss is independent of a previous toss and will always be ½.
Separate drawings from a deck of cards are independent events if you put the cards back.
Dependent events are the opposite. The probability of the second event is affected by the first event.
An example of a dependent event is drawing a card from a deck but not returning it. By not returning
the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of
whatever kind of card you drew. If you draw an ace of spades, there are 1 fewer aces and 1 fewer
spades. This affects our simple probability.
Note that dependent probabilities always coincide with “and” problems, so they will always be
multiplication problems.
Example 8
Two cards are pulled from a deck of 52 cards. They are pulled one after the other, and the first is not
returned to the deck. What is the probability that both cards will be spades?
Answer
Since both cards must be spades, this is an “and” question. We need to calculate the individual
probability of each card and then multiply.
The first card is like many of the other examples:
Bottom: 52 cards total
Top: 13 cards are spades
Probability: 13/52 = 1/4
For the second card, we have to pull it out after the first one has already been pulled. There are no
longer 52 cards, but rather 51. And, assuming the first card was a spade, there are now no longer 13
spades, but only 12. So the probability for the second card is:
Bottom: 51 cards left
Top: 12 cards are spades
Probability: 12/51=4/17
Finally, we multiply them together:
Probability of two spades: ¼ x 4/17 = 1/17
Note again that 1/17 is smaller than both ¼ and 4/17, since it is harder to pull out two spades in a row.
5
Example 9
There are 6 black marbles and 4 red marbles in a jar. Two marbles are pulled out in succession
without replacing them in the jar. What is the probability that both will be black?
Answer:
Notice that this is again a dependent probability situation. Once the first marble is pulled out, there are
less marbles for the second pull.
Probability the first marble will be black: 6/10
Assuming the first marble was black, there will now be only 5 black marbles left, and 9 marbles left all
together.
Probability the second marble will be black: 5/9
Total probability: 6/10 x 5/9 (don’t forget to reduce before you multiply) = 3/5 x 5/9 = 1/5 x 5/3 = 1/3
Strategice tip: This is where the GMAT starts getting tricky. The test assumes you have a good
understanding of these concepts, and takes all routes to test them. But don’t fear! While the
questions can be difficult, the GMAT does not go beyond this topic in probability. With the right
practice and a good deal of logic, you will be able to master these questions. Try the questions below
and see how they take the above concepts and tie them all together:
Example 10
After each throw of a red die, the face that shows is marked with a blue stripe. What is the probability
that after 6 throws all faces of the die will be marked red?
A.
B.
C.
D.
E.
Answer: B
First off, thinking through this problem logically is going to get you the right answer. It is a dependent
probability problem, because after each throw the number of sides the die can land on diminishes by 1.
Remember to think bottom to top for each step.
6
Throw One: At this point, the die can land on anything,
so there are 6 sides total and 6 sides that work for us, so
the probability is 6/6, or 1.
Throw Two: Now we have to assume that one of the
sides is marked in blue. We toss the die again, and we
hope it does not come up on the blue side. There are 5
sides that do give us what we want, out of 6 total, so the
probability is high: 5/6.
Throw Three: As in the previous throw, the number of
sides we want it to land on is reduced. Now there are
only 4 sides that it can land on, so the probability is 4/6.
Throw Four: By now you should see the pattern. If
we’ve been successful all along, then there are now three
sides with a blue mark, and three without. We want it to
land on one of the three that are not marked. The
probability is 3/6.
Throw Five: Now there are 4 sides marked, 2 not.
Probability of landing on one of those two: 2/6
Throw Six: This is the final throw. 5 sides should be
marked by now, with only one remaining. The
probability that we land on that one is 1/6.
We now have
6 5 4 3 2 1
      (remember to cancel)
6 6 6 6 6 6
5 1 1 1 1
    
3 3 2 3 6
5
(B)
324
Example 10
If a coin is tossed twice what is the probability that it will land either heads both times or tails both
times?
a)1/8
b)1/6
c)1/4
d)1/2
e)1
Answer: D
This question brings both “and” and “or” into the same problem. For the “or” part, we have two
scenarios that could both happen (both heads or both tails). But within each option, there are two
probabilities, and both must happen. (both must be heads or both must be tails). Let’s attack it
systematically:
Both Heads: This means heads on the first toss and heads on the second toss. Both tosses have a ½
probability of coming up heads, and ½ x ½ = ¼.
Both Tails: Same as heads, but reversed. Probability = ¼.
7
Both Heads or Both Tails: ¼ + ¼ = ½.
Example 11:
A fair coin is flipped three times. What is the probability that heads will come up only once?
Answer: 3/8
This is an “or” question, even though the “or” isn’t written. Why? Think about what it means to have
heads come up only once. What are the scenarios? Let’s write them out:
H T T or T H T or T T H
There are three situations, all of which are valid, that would achieve the result we’re looking for.
Again, let’s go one by one:
Probability of H T T = ½ x ½ x ½ = 1/8
Probability of T H T = ½ x ½ x ½ = 1/8
Probability of T T H = ½ x ½ x ½ = 1/8
Add 1/8 + 1/8 + 1/8 = 3/8
Example 12
The first jar contains 4 blue and 5 red marbles; the second basket contains 3 blue and 4 red marbles.
One marble is randomly extracted from the first basket and put into the second. After that, a marble is
extracted from the second basket. What is the probability that this marble is blue?
A.
B.
C.
D.
E.
1/3
15/36
31/72
4/9
11/18
Answer: C
Among all possible scenarios there are two that suit us:
1. A blue marble is put into the second basket and then a blue marble is extracted from the second
basket;
2. A red marble is put into the second basket and then a blue marble is extracted from the second
basket.
The probability of the first scenario: probability that a blue marble is taken from the first basket ×
probability that a blue marble is then extracted from the second basket = 4/9 × 1/2 = 4/18.
The probability of the second scenario: probability that a red marble is taken from the first basket ×
probability that a blue marble is then extracted from the second basket = 5/9 × 3/8 = 15/72.
The probability of EITHER first OR second scenario: 4/18 + 15/72 = 31/72.
8
The important thing is to understand that the probability of drawing a blue marble from the second
basket depends on what marble was put there. For example, in the first scenario the probability of
drawing a blue marble from the second basket is 1/2 because after one blue marble was added to the
second basket, the basket contains equal number of marbles of each color.
Example 13
In a 1 mile race, 5 athletes compete from each of three different schools: Washington High, Duke
High, and Cherry Hill High. What is the probability that a student from Cherry Hill will take first
place, a student from Duke will take second place, and another student from Duke will take third
place?
Answer: Multiply the Individual Probabilities
Probability that:
Cherry Hill student will take first: 5/15
Duke student takes second: 5/14
Duke student takes third: 4/13
Total Probability: 5/15 x 5/14 x 4/13 = 10/273
Example 14
Brian rolled two identical dice. What is the probability that the sum of the dice equaled 7?
Answer: 1/6
For this problem, like the others, we need to think very systematically. Let’s start by thinking about all
the situations that could yield a 7:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
As we saw in Example 7, each situation has a 1/36 chance of happening. Since any one of them would
yield 7, this is an “or” question. To solve, we’ll just add up all the probabilities: 1/36 + 1/36 + 1/36 +
1/36 + 1/36 + 1/36 = 6/36 = 1/6
IV. Working Backwards: Success, Failure, and the “At Least” Solution
Let’s go right back to the first example in this chapter:
Let’s say you have a single six-sided die. If you roll it, what is the probability you will roll a 1?
If you remember, the answer was 1/6.
Now let’s put some money on this. You’re in Vegas, and you’re going to win $500 if the die lands on
1. When it does, you jump up and down – you just won 500 dollars! Let’s call that a success. You
had a 1/6 chance to achieve that success.
But you decide to let it ride! You’re convinced it can happen again. You go double or nothing, betting
again on 1. This time, the die comes up on something else! You’ve failed. And since 5 out of the 6
outcomes would have caused that failure, the probability to fail is 5/6.
Now, let’s turn it around. What if someone were to offer you $500 if, when you roll one die, you
rolled at least a 2? Think about that for a second. What outcome would win you $500? If you’re just
thinking of 2, think again. In fact, if you rolled a 2, or a 3, or a 4, or a 5, or a 6, you would win! That’s
9
pretty good. What are the odds of that happening? You could figure out that there are 5 ways to win
out of 6, so the answer would be 5/6. The probability of success would be 5/6.
But, by using the phrase “at least,” the problem created more ways to succeed than to fail. That means
more work. On the GMAT, we want to avoid hard work. So what if you figure out how you could
fail, and then reverse it? The only way NOT to win $500 is to roll a 1. And what’s the probability that
you’d roll a 1? 1/6. The probability of failure is 1/6, so the probability of success must be 5/6.
To simplify it, look at this formula:
P(success) + P(failure) = 1
or
P(success) = 1 – P(failure)
Make sense? Let’s try another example.
Example 15
A jar contains 10 red marbles and 6 black marbles. If three marbles are pulled from the jar one after
another without replacing them back into the jar, what is the probability there will be at least one red
marble among them?
Answer: 27/28
Can you see why this would be tough? There are so many ways to pull out “at least one” red marble.
The red one can come first, second or third. Or, you could pull out two red marbles and one black one,
in any order, or you could pull out three red marbles. Figuring out the total number of ways to do that
would take a very long time.
But, think about the failure in this case. What is the one way to fail? You can fail by not pulling out
any reds at all! That’s right – in this case the failure is the case where you pull out three black
marbles!
So let’s figure that out. What is the probability of pulling out three black marbles? This is a dependent
probability problem, as you learned before. Just multiply the individual probabilities:
6/16x5/15x4/14 = 1/28
Since the probability of the failure, i.e., the probability of all black, is 1/28, the probability of the
success, i.e., the probability of at least one red, is 27/28.
Example 16
Two identical six-sided dice are rolled. What is the probability that the sum of the dice will be at least
5?
Answer: 5/6
Pay attention to the phrase “at least,” and think about what it means for this problem. “At least five”
means anything five or above. If we just counted all the ways the dice could come up at least five, it
would take several minutes. Rather, let’s figure out the failure, and reverse the answer. What is the
opposite of “at least five”? Anything less than five, or, in other words, up to four. So we have:
(1,1) = 2
(1,3) = 4
(2,2) = 4
10
(1,2) = 3
(2,1) = 3
(3,1) = 4
That’s all of them! So there are 6 ways to come up with 4 or less. As you have already seen, each one
has a probability of 1/36, so together the failure has a probability of 6/36 or 1/6.
The probability of success would then be 1 – 1/6 = 5/6.
V. A different method
NOTE: Please read this after you have mastered the other four sections.
You now have everything you need to succeed in GMAT probability questions. But to be fully
prepared for the test, you should be able to solve dependent and independent probabilities with just the
bottom to top approach. By utilizing what you know about permutations and combinations, you can
look at the whole probability question as a system, one in which you decide on the denominator and
the numerator separately, and then just put them together. Take a look at these examples:
Example 17
What is the probability that rolling two identical dice together will result in a 3 and a 4?
Answer: Think Bottom to Top
Bottom: We have two dice being rolled, with 36 total outcomes possible.
Top: There are two outcomes that can happen: (3,4) and (4,3).
Probability: 2/36 = 1/18
Example 18
What is the probability that rolling two identical dice together will result in the sum of the dice adding
to 7?
Answer: Think Bottom to Top
Note: We saw this question above in example 14. Try solving it a different way.
Bottom: Total outcomes = 36
Top: How many of those outcomes yield a 7? Just list them:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
There are 6 in total.
Probability: 6/36=1/6
Example 19
Bowl X has 4 cards in it, numbered 1-4. Bowl Y has 5 cards in it, numbered 5-9. What is the
probability that the sum of a card pulled randomly from Bowl X and a card pulled randomly from
Bowl Y will equal 8?
Answer: Think Bottom to Top
Bottom: How many different pairs of cards can be pulled? There are 4 cards in Bowl X and 5 cards in
Bowl Y. Using permutations: 4 x 5 = 20. So there are 20 total outcomes possible.
Top: Which pairs work for this question?
(1,7) (2,6) (3,5)
3 outcomes.
Probability: 3/20
11
Example 20
Bowl X has 4 cards in it, numbered 1-4. Bowl Y has 5 cards in it, numbered 5-9. What is the
probability that the product of a card pulled randomly from Bowl X and a card pulled randomly from
Bowl Y will be even?
Answer: Think Bottom to Top
Bottom: 20 total outcomes
Top: Which pairs will multiply to an even number?
(1,6)
(2,8)
(1,8)
(2,9)
(2,5)
(3,6)
(2,6)
(3,8)
(2,7)
(4,5)
(4,6)
(4,7)
(4,8)
(4,9)
14 total pairs.
Probability: 14/20=7/10
Alternative Answer
Think about success and failure. What is a success in this problem? An even outcome.
How can two numbers multiply together to make an even? There are lots of ways – all
we need is an even number. What if we reverse this one? Since there are only two
outcomes (odd and even), it should be clear that there are less odd pairs than even ones.
Odd outcomes are failures. Let’s calculate the probability of failure, and then subtract
from 1 to find the success:
(1,5)
(1,7)
(1,9)
(3,5)
(3,7)
(3,9)
Bottom: 20 total outcomes
Top: 6 odd outcomes
Probability of Odd Outcome (failure): 6/20 = 3/10
Probability of Even Outcome (success): 1 – 3/10 = 7/10
Example 21
A fair coin is flipped three times. What is the probability that heads will come up only
once?
Answer: Think Bottom to Top
Note: We saw this question above in example 11.
Bottom: How many different outcomes are possible? Each time the coin is tossed, there
are two outcomes, and we’re tossing it three times: 2 x 2 x 2 = 8
Top: How many outcomes have only a single instance of heads? There are only 8
possibilities, as we learned above, so there can’t be too many. Work it out:
HTT
THT
TTH
3 total outcomes
12
Probability: 3/8
VI. Extra Questions
1. A jar has 10 marbles, either black or white. 2 marbles are randomly chosen
simultaneously from the jar. If q is the probability that both will be black , is q >
1/3?
1) Less than ½ of the marbles in the jar are white.
2) The probability that 1 white marble and 1 black marble will be chosen
together is 7/15.
Ans: C
2. In a class with 12 children, q of the children are girls. Two children will be
randomly chosen simultaneously. What is the value of q?
1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is
16/33.
Ans: D
3. In a hotel with single rooms and double rooms, what is the probability that a room
chosen at random will be a double room painted red?
1) 1/6 of the rooms in the hotel are painted red.
2) 2/3 of the hotel’s rooms are double rooms.
Ans: E
4. Two identical dice are rolled together. If the sum of the dice is 7, what is the
probability that one of the numbers showing is a 4?
A) 1/36
B) 1/18
C) 1/6
D) 11/36
E) 1/3
Ans: E
5. At 3 pm, Jennifer went into labor. There is a 70% chance her baby will be born
each hour that she is in labor. What is the probability that her baby will be born at
6 pm on the same day?
A) .027
B) .063
C) .147
D) .27
E) .343
13
Ans: B
6. A fair coin is to be flipped four times. What is the probability that the coin will
land on the same side on all four flips?
A) 1/32
B) 1/16
C) 1/8
D) ¼
E) ½
Ans: C
14