# Probability

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This page is a crash-course to get you ready to crack probability questions on GMAT.

## Introduction

How probable is it to get probability questions on GMAT?
Probability questions are becoming increasingly common. They tend to be bundled among the difficult questions, so high scorers will commonly encounter 1, 2, or 3 of them. If you are a low scorer and are pressed for time, consider skipping this material, as you are not likely to encounter difficult questions on the adaptive test.
Do I have to be a genius to solve probability questions?
Absolutely not. Neither this brief course nor GMAT require any math knowledge beyond what you learned in your high school. Just put in some effort and you will crack it. Do not be influenced by the rumors that say probability is difficult - those come from a conspiracy among the instructors who want to secure their wages.
What is probability?
Probability is a measure of how likely is an event to happen. It is measured in fractions from 0 to 1 (0 is impossible, 1 is unavoidable or certain). Sometimes it is denoted in percentages, again from 0% to 100%.
What is an event and an outcome?
An event is anything that happens. In probability theory we speak of events having outcomes or results. A coin flip (an event) has two possible outcomes - heads and tails. A die toss has six possible outcomes. When a coin is flipped (an event is tested), one of the outcomes is obtained. Either heads or tails.
How is probability used?
The probability of event A is commonly denoted $p(A)$. So, if Heads is H and Tails is T, $p(H) = 0.5$ means that you have 1 chance in 2 to get heads in a coin flip. This also means that if you flip the coin 100 times, you will get heads about $100*0.5=50$ times. Not exactly 50. You may get 49, or 63, or even no heads. But 50 is the most likely number. This works, because we assume that the coin is fair, that coin flips are independent, and that we have accounted for all possible outcomes, heads and tails. This is always assumed on GMAT unless otherwise stated. But be sure to check these assumptions when using probability theory in business.

## Simple Probability: The F/T Rule

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes, when the outcoumes are equally likely. This is known as the F/T Rule, and 90% of the problems are solved using just this tool. No kidding.

Here are some examples to see how it works:

{{#x:box|

What is the probability that a card drawn at random from a deck of cards will be an ace?

Solution. In this case there are four favorable outcomes, aces of spades, hearts, diamonds and clubs.

Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes. Therefore, the probability is $\frac{4}{52}=\frac{1}{13}$. }}

The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice.

{{#x:box| Two fair six-sided dice are rolled; what is the probability of having 5 as the sum of the numbers?

Solution. There are 36 possible outcomes when a pair of dice is thrown (six outcomes for the first die times six outcomes for the second one). Since four of the outcomes have a total of 5, {(1,4), (4,1), (2,3), (3,2)}, the probability of the two numbers adding up to 5 is $\frac{4}{36} = \frac{1}{9}$. }}

## Probability of Multiple Events

For questions involving single events, the F/T rule is sufficient. In fact, it is often sufficient for all other cases too. But, for questions involving multiple events, some other tools may be more appropriate. Even when the problem can be solved with F/T, these tools still may provide a more elegant solution.

### NOT

If you know that the probability of an event (or one of the outcomes) is $p$, the probability of this event NOT happening (or the probability of it NOT having this given outcome), is $1-p$.

p(not A) + p(A) = 1


### AND

If two (or more) independent events are occurring, and you know the probability of each, the probability of BOTH (or ALL) of them occurring together (event A and event B and event C etc) is a multiplication of their probabilities.

p(A and B) = p(A) * p(B)
p(A and B and C ... and Z) = p(A) * p(B) * p(C) * ... * p(Z)


Suppose Mark will only be happy today if he gets an email and wins the lottery. He has a 90% chance to get an email and 0.1% chance to win the lottery. What are Mark's chances for happiness? Since email and lottery are independent (getting an email doesn't change my lottery chances, and vice versa), we can use the AND tool:

p(email AND lottery) = p(email) * p(lottery) = 90% * 0.1% = 0.09%


So Mark has 9 chances in 10,000. Not bad...

### OR

If two (or more) incompatible events are occurring, the probability of EITHER of them occurring (event A or event B or event C etc.) is a sum of their probabilities.

p(A or B) = p(A) + p(B)
p(A or B or C ... or Z) = p(A) + p(B) + ... + p(Z)


Incompatible means that they can't happen together, i.e. p(A and B) = 0. In case of two compatible events, the OR tool looks a bit more complicated:

p(A or B) = p(A) + p(B) - p(A and B)


If we know that A and B are independent, we can apply AND tool to rewrite:

p(A or B) = p(A) + p(B) - p(A) * p(B)


Suppose Mark will now be happy in both cases - either getting an email or winning the lottery. What are his chances to happiness now?

p(email OR lottery) = p(email) + p(lottery) - p(email) * p(lottery)


$90% + 0.1% - 0.09% = 90.01%$

Mark's chances are 9,001 in 10,000 now.

### Expressions/Brackets

When you're being asked for something complex, try reducing it to events and outcomes, and writing a formula. Use brackets to denote complex events, such as (A and B), or (A and (B or C)), etc. It is common to use AND as if it is multiplication and OR as if it is addition in the order preference, i.e. (A and B or C) = ((A and B) or C), but (A and (B or C)) <> (A and B or C). When you figure out the formula, it'll be easy to reduce it to simple arithmetic operations by using NOT, AND, and OR tools.

### Elimination tricks

Given that $0 \le p(A) \le 1$, you get the following rules:

p(A and B) $\le$ p(A)
p(A or B) $\ge$ p(A)
p(A and B) $\le$ p(A or B)


Thinking of these rules is often an excellent strategy for eliminating certain answer choices.

{{#x:box|

If a fair 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?

1. $\frac{1}{6}$
2. $\frac{1}{3}$
3. $\frac{1}{4}$
4. $\frac{1}{2}$
5. 1

Solution. Suppose first toss is A, second is B. We know that p(A_heads) = 50% and that p(B_tails) = 50%. Also, A and B are independent. So, p(A_heads and B_tails) = p(A_heads) * p(B_tails) = 50% * 50% = 25% = $\frac{1}{4}$. Answer is C. }}

{{#x:box| If a fair coin is tossed twice what is the probability that it will land either heads both times or tails both times?

1. $\frac{1}{8}$
2. $\frac{1}{6}$
3. $\frac{1}{4}$
4. $\frac{1}{2}$
5. 1

Solution. Let first toss be A, second B.

p(Ah) = p(At) = p(Bh) = p(Bt) = $\frac{1}{2}$

p(Ah and Bh) = p(Ah) * p(Bh) = $\frac{1}{4}$

p(At and Bt) = p(At) * p(Bt) = $\frac{1}{4}$

p((Ah and Bh) or (At and Bt)) = p(Ah and Bh) + p(At and Bt) = $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ }} Note that AND rule works because A and B are independent, and OR rule works because (Ah and Bh) and (At and Bt) are incompatible.

Alternatively, you may use F/T rule to solve this. Enumerate outcomes as (HH, HT, TH, TT). Favorable are HH and TT. So, p = $\frac{2}{4} = \frac{1}{2}$. Although in this case F/T rule works more gracefully, the AND/OR approach is still helpful - you can learn it on such easy examples as this to prepare for the more difficult ones.

{{#x:box| A bowman hits his target in $\frac{1}{2}$ of his shots. What is the probability of him missing the target at least once in three shots?

Solution. An optimal way to solve this is to think that (missing the target at least once) = 1 - (hitting it every time). So, p(hitting it every time) = p(shot1_hit and shot2_hit and shot3_hit) = p(shot1_hit) * p(shot2_hit) * p(shot3_hit) = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2} = \frac{1}{8}$; p(missing at least once) = 1 - p(hitting it every time) = $1 - \frac{1}{8} = \frac{7}{8}$. }} Alternatively, use the F/T rule. The T are HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM. T = 8. The F are HHM, HMH, HMM, MHH, MHM, MMH, MMM. F = 7.

In cases like this it is evident that F/T rule soon becomes too hard to apply.

## Event Types and Sets Analogy

### Compatible vs. Incompatible (Mutually exclusive) Events

Sometimes you have to distinguish compatible and mutually exclusive events. Mutually exclusive are those events that can't happen together. Heads and tails are mutually exclusive events. Formally, two events are mutually exclusive if p(A and B) = 0. Otherwise, they are compatible. Note that mutually exclusive events are independent. (!)

### Dependent vs. Independent Events

Most of the events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. 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 $\frac{1}{2}$. Separate drawings from a deck of cards are independent events if you put the cards back.

An example of a dependent event, one in which the probability of the second event is affected by the first, 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 one fewer aces and one fewer spades. This fact affects the F in the F/T rule.

What to do if you encounter dependent events? If possible, try to use F/T rule to the composite event of the two. In the cards example, you may consider counting all 2-card combinations you may draw (T), and then counting those that fit (F). This will be discussed in detail later. But sometimes the events can't be reduced to outcomes that can be counted. In these cases, use the sets analogy.

### Sets Analogy

Remember the familiar problem type about students attending three language classes, say, French, German, and Chinese? There you had to calculate the number of students attending one of the classes, or number of students attending both French and German, but not Chinese, etc? The greatest way to solve such problems is to draw intersecting circles representing the three sets of students, and then to write there their numbers and try to find the answer.

What does it have to do with probability, one might wonder. But this is precisely the way to solve probability problems with dependent events. This charts you may have drawn for simple sets problems are called Venn diagrams in the probability theory. Perhaps to scare you away.

The logic is simple: each event is a language class, and each chance is a student in that class. And the probability of the event is the number of students (chances) attending it divided by the total number of students. Where the classes intersect is where two events happen at once. Mutually exclusive events do not intersect. Finally, independent events intersect in such an interesting way that, supposing French and German classes represent two independent events, the proportion of French students in the German class is the same as the proportion of French students in the school as a whole (100 students, 40 study German, 50 study French, and 20 study both: $\frac{20}{40} = \frac{50}{100}$).

### Conditional Probability

Conditional probability is a simple way to denote proportions you understand with the sets analogy. Simply put, p(A/B) is the probability of event A happening given that event B has already happened, or the number of students attending both A and B classes divided by the number of students attending B class.

So, for any two events, including dependent events, this statement hold:

p(A and B) = p(A) * p(B/A) = p(B) * p(A/B)


This statement, however scary, is self-evident. Look at it. It says that to find the number of students studying French and German you have to either multiply the number of those who study French by the proportion of German scholars in the French class (p(B/A)), or multiply the number of German students by the proportion of French students in the German class (p(A/B)). But that's self-evident, isn't it? So it is with events.

Independent events may, therefore, be defined as such that p(B/A) = p(B), p(A/B) = p(A).

{{#x:box| What is the probability that a card selected from a deck will be either an ace or a spade?

1. $\frac{2}{52}$
2. $\frac{2}{13}$
3. $\frac{7}{26}$
4. $\frac{4}{13}$
5. $\frac{17}{52}$

Solution. Let A stand for a card being an ace, and S for it being a spade. We have to find p(A or S). Are A and S mutually exclusive? No. Are they independent? Why, yes, because spades have as many aces as any other suit. Then,

p(A or S) = p(A) + p(S) - p(A) * p(S)

With simple F/T we get:

p(A) = 4/52 = 1/13 p(S) = 13/52 = 1/4

So,

p(A or S) = 1/13 + 1/4 - 1/52 = 16/52 = 4/13 }}

Sets analogy can help you visualize the formula. Draw two intersecting circles - one for aces, the other for spades. To get the area (probability) of the figure formed by these two circles together (all chances that are either aces or spades), you add the areas of aces and spades and subtract the intersecting area, in order not to count it twice. What we subtract is the ace of spades that was counted twice.

Another way to think about the question is to just count aces and spades; that is, use the F/T rule. There are 13 spades in a deck and 3 aces other than the ace of spades already included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible outcomes, or $\frac{16}{52} = \frac{4}{13}$.

{{#x:box| If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?

Solution. Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = $\frac{4}{52} = \frac{1}{13}$. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B/A) = $\frac{3}{51} = \frac{1}{17}$, and therefore:

p(A and B) = p(A) * p(B/A) = $\frac{1}{13} * \frac{1}{17} = \frac{1}{221}$ }}

{{#x:box| If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw?

Solution. So, there are 12 red and 18 blue marbles. We are asked to draw twice and return the marble after each draw. Therefore, the first draw does not affect the probability of the second draw. We return the marble after the draw, and therefore, we return the situation to the initial conditions before the second draw. Nothing is altered in between draws; therefore, the events are independent.

p(drawing a red marble) would be $\frac{12}{30} = \frac{2}{5}$. The same is true for the second draw. Then p(First_Red and Second_Red) = p(First_Red) * p(Second_Red) = $\frac{2}{5} * \frac{2}{5} = \frac{4}{25}$. }}

{{#x:box| Now consider the same question with the condition that you do not return the marbles after each draw.

Solution. The probability of drawing a red marble on the first draw remains the same, $\frac{12}{30} = \frac{2}{5}$. The second draw, however, is different. The initial conditions have been altered by the first draw. We now have only 29 marbles in the jar and only 11 red. So, p(Second_Red/First_Red) = $\frac{11}{29}$. Using the dependent event formula,

p(First_Red and Second_Red) = p(First_Red) * p(Second_Red/First_Red) = $\frac{2}{5} * \frac{11}{29} = \frac{22}{145}$ }} To summarize, if you return every marble you select, the probability of drawing another marble is unaffected; the events are INDEPENDENT. If you do not return the marbles, the number of marbles is affected and therefore DEPENDENT.

## Learning the Advanced Tools

Detailed discussion of advanced solution tools is out of scope of this section, but here are some considerations to get you started:

### Combinations

Good understanding of CT formulas (n!, nAk, nCk) is essential to solving complex F/T problems, where both F and T are so large you can't enumerate them manually, but only with a formula. See our Combinations Lesson.

### Expectations

Some probability problems deal with money, gains, and bets. Often you have to calculate which bet will be better, or how much it will be worth. The tool that deals with this is Expectation. E = G * p, where G is gain, and p is probability. So, a 10% chance to get $100 is worth (has E) of$100 * 10% = $10. Therefore, it is better than to get$8 for granted, but worse than a 5% chance to get $300 (E =$300 * 5% = $15). Complex expectation works similarly: E1 = E * p, i.e. a 10% chance to get a 25% chance to get$100 is worth 10% * (25% * $100) =$2.5; This is how Expectations work.

### Distributions

The three types of distributions are Binominal, Hypergeometric, and Poisson distributions. These are just handy formulas for solving 3 very specific kinds of problems, like these:

• If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)
• There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)
• Each hour an average of ten cars arrive at the parking lot. The lot can handle at most fifteen cars per hour. What is the probability that at a given hour cars will not be accepted? (Poisson Distribution)

As you may have noticed, Poisson and Binominal Distribution problems are alike. In fact, these Distributions are two methods of solving the same kind of problems. The difference is that BD provides accurate but costly (many calculations) method, and PD provides and elegant approximation, and is therefore used only on large numbers.

While BD and HD are quite likely to appear on GMAT, PD is not. For GMAT Club's members it is an open question whether one can in fact encounter PD on GMAT. In any case, there won't be two questions on PD.

{{#x:box|Feel free to make necessary corrections to this guide --Dzyubam 13:56, 3 April 2008 (UTC) }}

## Other GMAT Areas of Interest

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