Skip to content
Combinatorics · Chapter 4

Probability

How likely is it? Count the outcomes you want, divide by all the outcomes, and multiply along the branches.

You'll need
open on YouTube
4.10%

Probability Fundamentals

In this section you will be able to find a probability by counting the outcomes you want and dividing by all the outcomes.

You roll two dice in a board game and need a 7 to win. Is that likely? Is a 7 easier to roll than a 2? Guess first!

Every roll of two dice is one cell in a 6 by 6 grid. All 36 cells are equally likely. So a probability is just a fraction: the cells you want, divided by all the cells.

Exploreplay first, formulas later

6 × 6 = 36 rolls; Sum is 7: 6/36 = 1/6

Die 1 (6) × Die 2 (6) = 36 equally likely rolls

123456123456234567345678456789567891067891011789101112Die 1 down the side, Die 2 across the top.
636=16\frac{6}{36} = \frac{1}{6}

Sum is 7: 6 of 36 rolls, probability 1/6

Name it

Definition 4.1.1 · Probability

Probability is the likelihood of something happening. To calculate probability, you need to know how many possible options or outcomes there are, and how many of them are the right ones.

Theorem 4.1.2 · Probability

probability=Total number of desired outcomesTotal number of possible outcomes\text{probability} = \frac{\text{Total number of desired outcomes}}{\text{Total number of possible outcomes}}

This works when every outcome is equally likely, like the faces of a fair die.

Watch itpredict, then reveal

Example 4.1

What is the probability of rolling a prime number on a 6 sided dice?

6 = 6 equally likely rolls

Die (6) = 6 equally likely rolls

123456123456Die.
Pick a condition to light up the cells.

6 = 6 equally likely rolls. Pick a condition.

  1. 1

    How many successful outcomes are there?

    Predict first:

Try it

Problem 4.7.1

Two 6-sided dice are rolled. What is the probability that the sum of the numbers rolled is 2?

Problem 4.7.2

Two 6-sided dice are rolled. What is the probability that the sum of the numbers rolled is 4?

Problem 4.7.3

Two dice are rolled. What is the probability that the sum of the numbers rolled is 6?

Contest cornerreal contest problems

Problem 4.7.7

AMC 8

The arrows on the two spinners shown below are spun. Let the number NN equal 10 times the number on Spinner A, added to the number on Spinner B. What is the probability that NN is a perfect square number?

Checkpoint for section 4.1

Get 2 of 3 right to master this section. One try each, no hints.
1

A bag has 3 red marbles and 5 blue marbles. You pull out one marble without looking. What is the probability it is red?

2

A whole number from 1 to 20 is picked at random. What is the probability that it is a multiple of 3?

3

Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

4.20%

Distinguishability

In this section you will be able to decide whether to count order, and avoid the trap of cases that are not equally likely.

You grab two cards from ten. Does it matter which card you grabbed first? For counting it might. For probability, often not, but there is one sneaky trap.

You may count with order or without order. Just be consistent: count the good outcomes and all the outcomes the same way. Look: if some cases can happen in more orders than others, they are not equally likely.

Exploreplay first, formulas later

Unordered: product is odd in 10 of 45 = 2/9

Pick 2 different cards from 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Light: the product is odd.

  • {1, 2}
  • {1, 3}
  • {1, 4}
  • {1, 5}
  • {1, 6}
  • {1, 7}
  • {1, 8}
  • {1, 9}
  • {1, 10}
  • {2, 3}
  • {2, 4}
  • {2, 5}
  • {2, 6}
  • {2, 7}
  • {2, 8}
  • {2, 9}
  • {2, 10}
  • {3, 4}
  • {3, 5}
  • {3, 6}
  • {3, 7}
  • {3, 8}
  • {3, 9}
  • {3, 10}
  • {4, 5}
  • {4, 6}
  • {4, 7}
  • {4, 8}
  • {4, 9}
  • {4, 10}
  • {5, 6}
  • {5, 7}
  • {5, 8}
  • {5, 9}
  • {5, 10}
  • {6, 7}
  • {6, 8}
  • {6, 9}
  • {6, 10}
  • {7, 8}
  • {7, 9}
  • {7, 10}
  • {8, 9}
  • {8, 10}
  • {9, 10}
1045=29\frac{10}{45} = \frac{2}{9}

Counting order instead gives 2090=29\frac{20}{90} = \frac{2}{9}. Same answer!

Order ignored: product is odd in 10 of 45 cases, 2/9

Name it

Remark 4.2.1

In probability problems, whether you decide to multiply for order or not is up to you, as both will give the same probability.

Remark 4.2.2

If you decide not to account for order when doing probability problems, make sure the number of orderings is the same for every case. In the first problem, every pair had 22 orderings, so it was okay to ignore order. In the second problem, the 22 cases had different numbers of orderings, so this didn't work.

Watch itpredict, then reveal

Example 4.2

I have cards numbered from 1 to 10. What is the probability I pick a pair of 2 different cards that have an odd product?

Unordered: product is odd in 10 of 45 = 2/9

Pick 2 different cards from 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Light: the product is odd.

  • {1, 2}
  • {1, 3}
  • {1, 4}
  • {1, 5}
  • {1, 6}
  • {1, 7}
  • {1, 8}
  • {1, 9}
  • {1, 10}
  • {2, 3}
  • {2, 4}
  • {2, 5}
  • {2, 6}
  • {2, 7}
  • {2, 8}
  • {2, 9}
  • {2, 10}
  • {3, 4}
  • {3, 5}
  • {3, 6}
  • {3, 7}
  • {3, 8}
  • {3, 9}
  • {3, 10}
  • {4, 5}
  • {4, 6}
  • {4, 7}
  • {4, 8}
  • {4, 9}
  • {4, 10}
  • {5, 6}
  • {5, 7}
  • {5, 8}
  • {5, 9}
  • {5, 10}
  • {6, 7}
  • {6, 8}
  • {6, 9}
  • {6, 10}
  • {7, 8}
  • {7, 9}
  • {7, 10}
  • {8, 9}
  • {8, 10}
  • {9, 10}
1045=29\frac{10}{45} = \frac{2}{9}

Counting order instead gives 2090=29\frac{20}{90} = \frac{2}{9}. Same answer!

Order ignored: product is odd in 10 of 45 cases, 2/9

  1. 1

    What must happen for the cards to have an odd product?

    Predict first:

Example 4.3

The sum of 2 positive integers is 4. Find the probability that one of the integers is a 2.

Unordered: one of them is a 2 in 1 of 2 = 1/2 (cases not equally likely; true answer 1/3)

Two positive integers add up to 4. Light: the one of them is a 2.

  • {1, 3}×2
  • {2, 2}×1
12\frac{1}{2}

Counting order instead gives 13\frac{1}{3}. Different answers!

Careful: these cases are not equally likely. Some stand for 2 ordered pairs, some for only 1. The true probability is 13\frac{1}{3}.

Order ignored: one of them is a 2 in 1 of 2 cases, 1/2. Careful: these cases are not equally likely

  1. 1

    What cases are there?

    Predict first:

Try it

Problem 4.7.16

AMC 10

Two different numbers are selected at random from {1,2,3,4,5}\{1, 2, 3, 4, 5\} and multiplied together. What is the probability that the product is even?

Contest cornerreal contest problems

Problem 4.7.6

AMC 8

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?

Problem 4.7.9

AMC 8

Two cards are dealt from a deck of four red cards labeled AA, BB, CC, DD and four green cards labeled AA, BB, CC, DD. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

Problem 4.7.18

AMC 8

Two different numbers are randomly selected from the set {−2,−1,0,3,4,5}\{-2, -1, 0, 3, 4, 5\} and multiplied together. What is the probability that the product is 0?

Checkpoint for section 4.2

Get 2 of 3 right to master this section. One try each, no hints.
1

Two different cards are picked from cards numbered 1 to 6. What is the probability that both cards are even?

2

You pick 2 different cards from cards numbered 1 to 6, and the order you pick them in matters. How many ordered pairs are there?

3

Two positive integers add up to 6. Every ordered pair, like (1, 5) and (5, 1), is equally likely. What is the probability that both integers are 3?

4.30%

Casework in Probability

In this section you will be able to split the good outcomes into cases, count each case, and add them up.

Roll three dice. That is 216 outcomes, far too many to list one by one. How can you count the good ones without going crazy?

Split the good outcomes into cases that don’t overlap. Count each case, then add. The total number of outcomes is usually the easy part.

Exploreplay first, formulas later

6 × 6 × 6 = 216 rolls; Sum is 18: 1/216 = 1/216

Die 1 (6) × Die 2 (6) × Die 3 (6) = 216 equally likely rolls

Die 3 = 1
1234561234563456784567895678910678910117891011128910111213Die 1 down the side, Die 2 across the top.
Die 3 = 2
123456123456456789567891067891011789101112891011121391011121314Die 1 down the side, Die 2 across the top.
Die 3 = 3
123456123456567891067891011789101112891011121391011121314101112131415Die 1 down the side, Die 2 across the top.
Die 3 = 4
12345612345667891011789101112891011121391011121314101112131415111213141516Die 1 down the side, Die 2 across the top.
Die 3 = 5
123456123456789101112891011121391011121314101112131415111213141516121314151617Die 1 down the side, Die 2 across the top.
Die 3 = 6
123456123456891011121391011121314101112131415111213141516121314151617131415161718Die 1 down the side, Die 2 across the top.
1216\frac{1}{216}

Sum is 18: 1 of 216 rolls, probability 1/216

Name it

Concept 4.3.1 · Casework in Probability

Often, when a problem asks for the probability of an event, you may have to do casework to find the number of successful outcomes. Then divide by the total number of outcomes, which is usually easy to find.

Watch itpredict, then reveal

Example 4.4

What is the probability that the product of 4 dice rolls is a multiple of 648?

  1. 1

    Break 648648 into primes. How many factors of 33 does it have?

    Predict first:
open on YouTube

Try it

Problem 4.7.5

AMC 8 Modified

A fair 6 sided die is rolled twice. What is the probability that the first number that comes up is greater than the second number?

Problem 4.7.17

MATHCOUNTS

A fair tetrahedral die, whose faces are numbered 1, 2, 3, and 4 is rolled three times. What is the probability that the sum of the numbers rolled is 7? Express your answer as a common fraction.

Contest cornerreal contest problems

Problem 4.7.10

AMC 10

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

Problem 4.7.11

AMC 8

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

Problem 4.7.12

AMC 10

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as

n67,\frac{n}{6^7},

where nn is a positive integer. What is nn?

Checkpoint for section 4.3

Get 2 of 3 right to master this section. One try each, no hints.
1

Two 6-sided dice are rolled. What is the probability that the sum is at least 10?

2

Two 6-sided dice are rolled. What is the probability that the product is a multiple of 5?

3

A fair coin is tossed 3 times. What is the probability of getting exactly 2 heads?

4.40%

Probability of Independent Events

In this section you will be able to multiply the probabilities of independent events.

You flip a coin twice. The coin has no memory: the first flip can’t change the second. So what is the chance of two heads?

Two events are independent when one doesn’t change the chances of the other. Then you can find each probability on its own and multiply. A tree shows why: each branch splits again by the same fractions.

Exploreplay first, formulas later

Leaves 1/4, 1/4, 1/4, 1/4; highlighted sum 1/4

P(1st heads)1/2
P(2nd heads | 1st heads)1/2
Coin 1Coin 21/2Heads1/2Heads1/2Tails1/2Tails1/2Heads1/2Tails
Highlighted total: 1/4

Independent? P(2nd heads | 1st heads) = 1/2, and P(2nd heads) = 1/2. Equal, so yes: independent.

Highlighted: Heads then Heads. Total 1/4. Independent.

Name it

Remark 4.4.1

Whenever you need the probability of 22 independent events both happening (the results of the events don't depend on each other), find the probability of each event on its own and multiply them together.

Watch itpredict, then reveal

Example 4.5

Sohil randomly picks a number from 1 to 10. Sejal randomly picks a number from 1 to 25. What is the probability that the product of the numbers they choose is odd?

Leaves 13/50, 6/25, 13/50, 6/25; no leaves highlighted

P(Sohil odd)5/10
P(Sejal odd | Sohil odd)13/25
SohilSejal1/2Sohil odd13/25Sejal odd12/25Sejal even1/2Sohil even13/25Sejal odd12/25Sejal even
Click leaves to add them up

Independent? P(Sejal odd | Sohil odd) = 13/25, and P(Sejal odd) = 13/25. Equal, so yes: independent.

Highlighted: none. Total 0. Independent.

  1. 1

    To start, what do you know about 2 numbers whose product is odd?

    Predict first:

Example 4.6

Omega Learn

Person A rolls a dice with the numbers 1, 1, 2, 6, 15, 30 and Person B rolls a dice with the numbers 1, 2, 4, 18, 28, 44. What is the probability that the product of the 2 numbers is a multiple of 24?

6 × 6 = 36 equally likely rolls

Person A (6) × Person B (6) = 36 equally likely rolls

12418284411261530Person A down the side, Person B across the top.
Pick a condition to light up the cells.

6 × 6 = 36 equally likely rolls. Pick a condition.

  1. 1

    How many outcomes are there in total?

    Predict first:
open on YouTube

Try it

Problem 4.7.4

AMC 8

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

Contest cornerreal contest problems

Problem 4.7.19

Omega Learn Math Competition

Joe is standing at the point (0,0) in the infinite coordinate plane. On any given move, there is a 13\frac{1}{3} chance of him moving up, 13\frac{1}{3} chance of him moving right, 16\frac{1}{6} chance of him moving down, and 16\frac{1}{6} chance of him moving left. If the probability he will reach the point (3,3) in less than or equal to 8 moves can be described as a3b\frac{a}{3^b}, then what is the value of a+ba + b?

Problem 4.7.20

Omega Learn Math Competition

The Blazers and Cortzans are playing a series of 7 games. A team wins the series when they have won 4 games. Both teams have an equal likelihood of winning a game. If the probability the Cortzans win the series and lose less than 3 games can be expressed as mn\frac{m}{n}, find the value of m+nm + n?

Problem 4.7.21

AMC 8

On the dart board shown in the figure below, the outer circle has radius 6 and the inner circle has radius 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?

Checkpoint for section 4.4

Get 2 of 3 right to master this section. One try each, no hints.
1

You flip a coin and roll a 6-sided die. What is the probability of getting heads and a 6?

2

You spin a spinner numbered 1 to 4 and roll a 6-sided die. What is the probability that both numbers are even?

3

A fair coin is tossed 4 times. What is the probability that every toss is heads?

4.50%

Probability of Dependent Events

In this section you will be able to multiply along the branches when the second event depends on the first.

Seven runners race. If Chase wins, can George still come 1st? No! Knowing one result changes the chances of the other.

When events are dependent, multiply the first probability by the second probability given that the first one happened. On a tree, that means: multiply along a path, then add the paths you want.

Exploreplay first, formulas later

Leaves 6%, 14%, 8%, 72%; no leaves highlighted

P(rain Sat)20%
P(rain Sun | rain Sat)30%
P(rain Sun | no rain Sat)10%
SaturdaySunday20%Rain30%Rain70%No rain80%No rain10%Rain90%No rain
Click leaves to add them up

Independent? P(rain Sun | rain Sat) = 30%, and P(rain Sun) = 14%. Not equal, so no: dependent.

Highlighted: none. Total 0%. Dependent.

Name it

Remark 4.5.1

When the events are dependent, like in Example 4.7, the probability of both events AA and BB happening is the probability of AA times the probability that BB happens given that AA already happened:

P(A and B)=P(A)×P(B∣A).P(A \text{ and } B) = P(A) \times P(B \mid A).

Watch itpredict, then reveal

Example 4.7

Alex, Betty, Chase, Derek, Emma, Fiona, and George are racing in a marathon. If they finish in a random order, what is the probability that Chase is 1st and George is 6th?

Leaves 1/42, 5/42, 5/42, 31/42; no leaves highlighted

1st place6th place1/7Chase 1st1/6George 6th5/6George not 6th6/7Chase not 1st5/36George 6th31/36George not 6th
Click leaves to add them up

Independent? P(George 6th | Chase 1st) = 1/6, and P(George 6th) = 1/7. Not equal, so no: dependent.

Highlighted: none. Total 0. Dependent.

  1. 1

    Can you just multiply P(Chase 1st) by P(George 6th) like in Section 4.4?

    Predict first:

Example 4.8

On Saturday, there is a 20% chance of rain. On Sunday, there is a 30% chance of rain if it rained on Saturday, but only a 10% chance of rain if it didn't rain on Saturday. What is the probability that it will rain on both days or neither day?

Leaves 6%, 14%, 8%, 72%; no leaves highlighted

P(rain Sat)20%
P(rain Sun | rain Sat)30%
P(rain Sun | no rain Sat)10%
SaturdaySunday20%Rain30%Rain70%No rain80%No rain10%Rain90%No rain
Click leaves to add them up

Independent? P(rain Sun | rain Sat) = 30%, and P(rain Sun) = 14%. Not equal, so no: dependent.

Highlighted: none. Total 0%. Dependent.

  1. 1

    What is the probability that it rains on both days?

    Predict first:
open on YouTube

Example 4.9

AIME

There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is ab\frac{a}{b}, where aa and bb are relatively prime positive integers. Find a+ba + b.

Leaves 0, 2/5, 0, 3/5; no leaves highlighted

P(rain Sat)4/10
P(rain Sun | rain Sat)0/14
P(rain Sun | no rain Sat)0/14
SaturdaySunday2/5Rain0Rain1No rain3/5No rain0Rain1No rain
Click leaves to add them up

Independent? P(rain Sun | rain Sat) = 0, and P(rain Sun) = 0. Equal, so yes: independent.

Highlighted: none. Total 0. Independent.

  1. 1

    Let xx be the chance of rain on Sunday if it does not rain Saturday. What is the chance of rain on Sunday if it does rain Saturday?

open on YouTube

Try it

Problem 4.7.15

Omega Learn Math Contest

John has 9 red apples and 3 green apples. He randomly selects and eats an apple. Then, 3 of the red apples get rotten and are thrown away. From the remaining apples he randomly selects and eats another apple. What the probability that both of the selected apples were red is ab\frac{a}{b}, what is the value of a+ba + b?

Contest cornerreal contest problems

Problem 4.7.8

AMC 8

A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

Problem 4.7.13

AMC 10

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?

Problem 4.7.14

AMC 8

On a beach 50 people are wearing sunglasses and 35 people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is 25\frac{2}{5}. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

Problem 4.7.22

Omega Learn Math Competition

Eric is stacking twenty six Jenga Blocks. For n≥1n \ge 1, the probability that placing the nnth block causes the whole tower to topple is 126−n\frac{1}{26-n} assuming the tower still stands after placing n−1n - 1 blocks. What is the expected number of blocks placed before the block that causes the tower to topple?

Checkpoint for section 4.5

Get 2 of 3 right to master this section. One try each, no hints.
1

A bag has 4 red and 2 blue marbles. You draw two marbles without putting the first one back. What is the probability that both are red?

2

Ann, Ben, Cat, Dan and Eve finish a race in a random order. What is the probability that Ann is 1st and Ben is 2nd?

3

There is a 50% chance of rain on Monday. If it rains Monday, there is a 60% chance of rain on Tuesday; if not, only a 20% chance. What is the probability it rains on both days?

4.60%

Dependent or Independent?

In this section you will be able to tell dependent from independent events, and handle both.

Mark and Steve are enemies, and the teacher picks seats at random. Will Steve end up in Mark’s row or column?

Ask yourself: does the first result change the choices for the second? Here Mark takes a chair, so Steve has one chair fewer. That makes the seats dependent. Count what is left after the first choice.

Exploreplay first, formulas later

3 × 3 chairs: Steve has 8 seats, 4 not in Mark's row or column, probability 1/2

rows3
columns3

Click a chair to seat Mark. Shaded chairs share his row or column.

48=12\frac{4}{8} = \frac{1}{2}

Steve has 8 open chairs: 4 in Mark's row or column, 4 safe. Move Mark around: does the answer change?

Mark sits in row 2, column 2. Steve has 8 seats; 4 share Mark's row or column, 4 do not. Probability 1/2.

Watch itpredict, then reveal

Example 4.10

Omega Learn

Mark must choose a seat from 5×55 \times 5 grid of chairs in the classroom. His enemy Steve does not like Mark so wants to choose a seat that is not in the same row or column as Mark. However, to his dismay, the teacher randomly assigns seats. What is the probability that Steve is not sitting in the same row or column as Mark?

5 × 5 chairs: Steve has 24 seats, 16 not in Mark's row or column, probability 2/3

Click a chair to seat Mark. Shaded chairs share his row or column.

Probability: count, then divide!

Steve has 24 open chairs: 8 in Mark's row or column, 16 safe. Move Mark around: does the answer change?

Mark sits in row 3, column 3. Steve has 24 seats; 8 share Mark's row or column, 16 do not.

  1. 1

    Put Mark in any chair. How many chairs are left for Steve?

    Predict first:
open on YouTube

Contest cornerreal contest problems

Problem 4.7.25(additional)

MATHCOUNTS

Consider the regular octahedron as shown. Each edge of the octahedron has a length of 1. An ant starts at the vertex A and crawls a total distance of 3 units along the edges of the octahedron. Any time the ant reaches a vertex of the octahedron, it randomly chooses an edge to next crawl on that is different from the edge it just left. One such path the ant may take is shown. What is the probability that the ant will end up a distance greater than 1 from its starting point A ? Express your answer as a common fraction.

Checkpoint for section 4.6

Get 2 of 3 right to master this section. One try each, no hints.
1

Mark and Steve are randomly given two different seats in a 4 × 4 grid of chairs. What is the probability that Steve is not in Mark's row or column?

2

Two cards are drawn without replacement from a standard 52-card deck. What is the probability that both are aces?

3

You roll a die twice. Are the events "the first roll is 6" and "the second roll is 6" independent?

4.7

Practice Problems

In this section you will be able to mix every probability trick from this chapter on contest problems.

Time to mix it all up! These problems use counting, casework, ordered pairs and trees. Every problem has three hints if you get stuck.

Want every problem from this chapter on one page? Open the problem set.

Mixed practice

Problem 4.7.23(additional)

AMC 8

Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

Contest cornerreal contest problems

Problem 4.7.24(additional)

MATHCOUNTS

A positive integer divisor of 11!11! is chosen at random. What is the probability that this divisor is prime? Express your answer as a common fraction.