Often, finding the number of possible outcomes is the main component in solving a probability problem. The following examples show different methods for finding the total number of outcomes.

## Tree Diagrams

## Example

Two fair coins are tossed and a six-sided die is rolled. What is the probability of getting 2 heads and a 6?

### Solution

There is only one desired outcome:Â **H H 6**

You need to find the total number of possible outcomes.

Assume that the coin tosses and rolling the die happen one at a time.

*Method 1*

One method is to visualize possible outcomes with a tree diagram.

The first toss has 2 possible outcomes.

The second toss has 2 possible outcomes for each of the first outcomes.

The roll of the die has 6 possible outcomes for each of the outcomes from tossing the coins.

There are a total of 24 possible outcomes. The probability is \dfrac{1}{24}.

*Method 2*

Another method is multiplying the number of possible outcomes for each action.

The first toss has 2 possible outcomes.

The second toss has 2 possible outcomes.

Rolling the die has 6 possible outcomes.

The total of possible outcomes is

2 Ã— 2 Ã— 6 = 24.

The probability is \dfrac{1}{24}

## Outcomes

Another method is to use the number of outcomes rather than multiplying each probability.

## Example

A fair coin is to be tossed four times. What is the probability that the coin will land on the same side on all four tosses?

(A) \dfrac{1}{32}

(B) \dfrac{1}{16}

(C) \dfrac{1}{8}

(D) \dfrac{1}{4}

(E) \dfrac{1}{2}

### Solution

*Method 1**Â Â Â Using number of outcomes*

There are 2 possible outcomes for each of the 4 tosses. So the total number of outcomes is

2 Ã— 2 Ã— 2 Ã— 2 = 16.

There are 2 possible desired outcomes:Â **H H H H**Â orÂ **T T T T**

Using the outcomes, the probability is \dfrac{2}{16} = \dfrac{1}{8}.

The correct answer is choice (C).

*Method 2**Â Â Â Using probabilities*

The probability that the coin will land on the same side is the probability that it will land on all heads or all tails.

The probability of for each toss is \dfrac{1}{2} Ã— \dfrac{1}{2} Ã— \dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{16}. Since either scenario would satisfy the condition, add the two probabilities together.

\dfrac{1}{16} + \dfrac{1}{16} = \dfrac{2}{16} = \dfrac{1}{8}.

The correct answer is choice (C).

## Example

You roll 3 dice. What is the probability that:

a) all 3 dice match?

### Solution

*Method 1**Â Â Â Using number of outcomes*

Rolling 3 dice has 6^{3}Â = 216 possible outcomes. You can find this by multiplying the number of possible outcomes for each roll, using a tree diagram or using combinations (see the chapter Permutations and Combinations).

Assume that rolling the dice happens one die at a time.

**a)
**Rolling the first die has 6 possible outcomes.

There are 5 outcomes that do not match the first value. There are 4 outcomes that do not match either of the first 2 values.

The probability none of the dice match is \dfrac{5}{9}.

*Method 2**Â Â Â Using probabilities*

Assume that rolling the dice happens one die at a time.

**a)**

Any value can be the first value, so it does not affect the probability.

The probability that the second roll does not match the first is 5 values out of 6 possible values, so \dfrac{5}{6}.

The probability that the third roll does not match either of the two first rolls is 4 values out of 6 possible values, so \dfrac{4}{6}.

1 Ã— \dfrac{5}{6 }Ã— \dfrac{4}{6} = \dfrac{5 Ã— 4}{2 Ã— 3 Ã— 2 Ã— 3} = \dfrac{5}{9}

The probability none of the dice match is \dfrac{5}{9}.

## Example

You roll 3 dice. What is the probability that:

b) the sum of the dice will not be 6?

### Solution

*Method 1**Â Â Â Using number of outcomes*

Rolling 3 dice has 6^{3}Â = 216 possible outcomes. You can find this by multiplying the number of possible outcomes for each roll, using a tree diagram or using combinations (see the chapter Permutations and Combinations).

Assume that rolling the dice happens one die at a time.

**b)**

Rolling the first die has 6 possible outcomes.

There is 1 outcome that matches the first value for each of the other 2 dice.

The probability all 3 of the dice match is \dfrac{1}{36}.

*Method 2**Â Â Â Using probabilities*

Assume that rolling the dice happens one die at a time.

**b)**

Any value can be the first value, so it does not affect the probability.

The probability that the second roll matches the first is 1 value out of 6 possible values, so \dfrac{1}{6}.

The probability that the third roll matches the two first rolls is also 1 value out of 6 possible values, so \dfrac{1}{6}.

1 Ã— \dfrac{1}{6} Ã— \dfrac{1}{6} = \dfrac{1}{36}

The probability all 3 of the dice match is \dfrac{1}{36}.

## Example

You roll 3 dice. What is the probability that:

c) exactly 2 of the dice match?

### Solution

*Method 1**Â Â Â Using number of outcomes*

Rolling 3 dice has 6^{3}Â = 216 possible outcomes. You can find this by multiplying the number of possible outcomes for each roll, using a tree diagram or using combinations (see the chapter Permutations and Combinations).

Assume that rolling the dice happens one die at a time.

**c)**

There are 3 pairings where the values can match:

first and second

first and third

second and third

So we will calculate the probability of one of these pairs occurring, and multiply that by 3 to get the total probability that any one of the three will occur.

Rolling the first die has 6 possible outcomes.

For the remaining two dice, one value must match and one value must not match that first die.

The probability that one pair of the dice match is \dfrac{5}{36}

So the probability that any of the 3 pairings of the dice match is:

3(\dfrac{5}{36}) = \dfrac{3 Ã— 5}{3 Ã— 12} = \dfrac{5}{12}

*Method 2**Â Â Â Using probabilities*

Assume that rolling the dice happens one die at a time.

**c)**

There are 3 pairings where the values can match:

first and second

first and third

second and third

One die can have any value, so it doesn’t affect the probability.

Of the remaining two dice, one must match and one must not match that first die.

The probability of a match is \dfrac{1}{6}. The probability of not matching is \dfrac{5}{6}.

1 Ã— \dfrac{1}{6} Ã— \dfrac{5}{6} = \dfrac{5}{36}

The probability that any single pair of the dice match is \dfrac{5}{36}.

There are 3 possible pairings.

So the probability of having exactly 2 dice match is 3(\dfrac{5}{36}) = \dfrac{5}{12}.

**Note:** For (c), there is another method. If figuring out the options for the pairings is not clear to you, you can calculate the probabilities for (a) and (b) and subtract from 1.

probability that exactly 2 match

= 1 â€“ (probability that none match or 3 match)

probability that exactly 2 match

= 1 â€“ ( \dfrac{5}{9} + \dfrac{1}{36} ) = 1 \,-\, ( \dfrac{20}{36} + \dfrac{1}{36} )

= \dfrac{36}{36} \,-\, \dfrac{21}{36} = \dfrac{15}{36} = \dfrac{5}{12}

## Lists or Tables

For scenarios that have few possible outcomes, making a list of the desired outcomes can be the simplest.

Video Courtesy of (__site-affiliate__) Kaplan GMAT prep. $200 off Kaplan Tutoring

## Example

Two six-sided 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) \dfrac{1}{36}

(B) \dfrac{1}{18}

(C) \dfrac{1}{6}

(D) \dfrac{11}{36}

(E) \dfrac{1}{3}

### Solution

You want to calculate the probability of rolling a 4 knowing that the sum of the dice is 7. Therefore the total possible outcomes are just the pairs of dice that sum to 7.

(1, 6)Â Â Â (2, 5)Â Â Â (3, 4)

(6, 1)Â Â Â (5, 2)Â Â Â (4, 3)

There are 6 pairs that sum to 7, and 2 of the pairs have a 4 in them. The probability is \dfrac{2}{6} = \dfrac{1}{3}.

The correct answer is choice (E).

## Example

Bowl A contains 4 cards, each numbered 1 to 4. Bowl B has 5 cards, numbered 5 to 9. What is the probability that the sum of a card pulled randomly from Bowl A and a card pulled randomly from Bowl B will equal 8?

### Solution

How many different pairs of cards can be pulled?

There are 4 cards in Bowl A and 5 cards in Bowl B. Multiplying the number of outcomes: 4 Ã— 5 = 20. So there are 20 total possible outcomes.

Which pairs add to 8? For this small number of options, a list is the easiest.

(1, 7), (2, 6), (3, 5)

So the probability is \dfrac{3}{20}.

## Example

Bowl A contains 4 cards, each numbered 1 to 4. Bowl B has 5 cards, numbered 5 to 9. What is the probability that the product of a card pulled randomly from Bowl A and a card pulled randomly from Bowl B will be even?

### Solution

There are 4 cards in Bowl A and 5 cards in Bowl B. Multiplying the number of outcomes: 4 Ã— 5 = 20. So there are 20 total possible outcomes.

Which pairs multiply to be even? An even number just needs one even factor, so an even card in one bowl multiplied by any card in the other bowl will be even.

*Method 1**Â Â Â Using outcomes*

Bowl A has 2 even cards and Bowl B has 5 cards, so that creates 10 even outcomes.

Bowl B has 2 even cards and Bowl A has 4 cards, so that creates 8 even outcomes.

Bowl A has 2 even cards and Bowl B has 2 even cards, so the number of outcomes that will have both cards even is 4. These outcomes are being counted twice, so subtract them.

So the number of even outcomes is 10 + 8 â€“ 4 = 14.

The probability is \dfrac{14}{20} = \dfrac{7}{10}.

*Method 2**Â Â Â Use a list of desired outcomes*

Make a list of possible pairings. Write all the numbers from Box A with the numbers from Box B that make even products.

(1, 6) | (2, 5) | (3, 6) | (4, 5) |

(1, 8) | (2, 6) | (3, 8) | (4, 6) |

(2, 7) | (4, 7) | ||

(2, 8) | (4, 8) | ||

(2, 9) | (4, 9) |

So there are 14 pairs that have even outcomes. The probability is

\dfrac{14}{20} = \dfrac{7}{10}.

*Method 3**Â Â Â Use a list of failed outcomes*

Look for the pairs that have odd outcomes. Odd outcomes are not desired, so (1 â€“ odd outcomes) is the probability of an even outcome.

(1, 5) | (3, 5) |

(1, 7) | (3, 7) |

(1, 9) | (3, 9) |

So there are 6 pairs that have odd outcomes. The probability is

1 â€“ (\dfrac{6}{20}) = \dfrac{14}{20} = \dfrac{7}{10}.

## Data Sufficiency

Probability is a common topic for Data Sufficiency questions.

ExampleA jar has 10 marbles, a mix of red and white. Two marbles are randomly chosen from the jar. If b is the probability that both will be red, is b > \dfrac{1}{3}?

(1) Less than \dfrac{1}{2} of the marbles in the jar are white.

(2) The probability that 1 white marble and 1 red marble will be chosen together is \dfrac{7}{15}.A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

### Solution

This is a dependent probability problem. To find the probability of choosing 2 red marbles, you need to figure out the probability that the first marble will be red and that the second marble will be red. In this case, the question wants to know if that probability is larger than \dfrac{1}{3}.

Statement (1) says that less than half the marbles are white, which means that more than half the marbles are red. The best way to approach this is to systematically (but quickly) figure out what the probability of drawing two red marbles is for each scenario. Make a table where

(number of red) > (number of white) and the numbers sum to 10.

Number ofred marbles |
Number ofwhite marbles |
Probability of2 red marbles |

6 | 4 | \dfrac{6}{10} Ã— \dfrac{5}{9} = \dfrac{1}{3} |

7 | 3 | \dfrac{7}{10} Ã— \dfrac{6}{9} = \dfrac{7}{15} |

8 | 2 | \dfrac{8}{10} Ã— \dfrac{7}{9} = \dfrac{28}{45} |

9 | 1 | \dfrac{9}{10} Ã— \dfrac{8}{9} = \dfrac{4}{5} |

10 | 0 | \dfrac{10}{10} Ã— \dfrac{9}{9} = 1 |

When less than half the marbles are white, the probability of choosing 2 red marbles can be greater than or equal to \dfrac{1}{3}. Statement (1) is not sufficient.

Statement (2) says the probability of choosing one red marble and one white marble is \dfrac{7}{15}. This is a trap. Since the probability given is exact, it may seem that only one scenario of red marbles and white marbles will work.

If you make a table of all the scenarios, you will see that when there are 7 red marbles and 3 white marbles, the probability of choosing one of each is \dfrac{7}{15}. However, it is also true in reverse: If there were 7 white marbles and 3 red marbles, the probability would also be \dfrac{7}{15}. Therefore, Statement (2) is not sufficient.

Combining Statements (1) and (2)Â **does**Â give enough information. From Statement (2) you know that there must be 7 of one color and 3 of the other. From Statement (1) you know that there must be more red than white. The only combination that fits this is 7 red marbles and 3 white marbles.

The correct answer is choice (C).