Unit 2 · Topic 2.4 Beta

Introduction to Probability

List the sample space.

Practice 3: Analyze Data

Part 1 · Hook

Why this matters

A board game says you escape jail by rolling doubles. Your friend rolls a sum of 7 three turns in a row and swears the dice are loaded. Are some rolls really more common? List the 36 ways two dice can land, and the answers come from counting.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. "At least 3" heads in 5 flips means

  1. 3, 4 or 5 heads
  2. 0, 1, 2 or 3 heads
  3. exactly 3 heads
  4. 4 or 5 heads
Show the answer

At least 3 means 3 or more.

  • Correct: 3, 4 or 5 heads:
  • 0, 1, 2 or 3 heads:
  • exactly 3 heads:
  • 4 or 5 heads:

2. In 500 simulated trials, a result happened 85 times. What is the estimated probability?

  1. 0.17
  2. 0.85
  3. 85
  4. 0.085
Show the answer

85500=0.17\frac{85}{500} = 0.17: trials with the result over all trials.

  • Correct: 0.17:
  • 0.85:
  • 85:
  • 0.085:

Part 4 · See it

See it first

A 6 by 6 grid of the 36 equally likely outcomes when a red die and a blue die are rolled. The 5 outcomes with a sum of 8 are shaded: (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2). So P(sum is 8) = 5/36.
The 36 equally likely outcomes of two dice. Five of them give a sum of 8. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. List every possible outcome of the chance processthe sample space (36 pairs for two dice)
  2. The outcomes are equally likelythe probability of an event is the number of its outcomes over the total: P(sum is 8)=536P(\text{sum is } 8) = \frac{5}{36}
  3. An event is easier to count from the outsideuse the complement rule: P(A)=1−P(Ac)P(A) = 1 - P(A^c)
  4. The outcomes are not equally likely (win, lose, tie)counting outcomes fails; use the long-run shares instead

Part 6 · Key ideas

Key ideas

  • The sample space is the list of every possible outcome. An event is a set of outcomes, such as "sum is 8".
  • When outcomes are equally likely, P(A)P(A)=number of outcomes in Anumber of outcomes= \frac{\text{number of outcomes in } A}{\text{number of outcomes}}.
  • Every probability is from 0 to 1, and the probabilities of all the outcomes add to 1.
  • The complement AcA^c is "A does not happen": P(Ac)=1−P(A)P(A^c) = 1 - P(A). It is the quick route for "at least one".

Part 7 · Misconception

A common mistake

The wrong idea: A sum of 12 has probability 111\frac{1}{11}, since there are 11 possible sums.

What actually happens: The sums are not equally likely. Count the 36 equally likely pairs: only (6, 6) gives 12, so the probability is 136\frac{1}{36}.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Rolling a red die and a blue die

A red die and a blue die, both fair, are rolled. The grid lists the 36 equally likely outcomes as (red, blue). The shaded cells are the outcomes with a sum of 8.
A 6 by 6 grid of the 36 equally likely outcomes when a red die and a blue die are rolled. The 5 outcomes with a sum of 8 are shaded: (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2). So P(sum is 8) = 5/36.
LevlPrep original diagram.

1. What is the probability that the sum of the two dice is 7?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Six outcomes give a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). With 36 equally likely outcomes, P(sum is 7)P(\text{sum is } 7)=636= \frac{6}{36}≈0.167\approx 0.167.

  • Answer: 0.1667

2. What is the probability that the sum is at least 10?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

"At least 10" means 10, 11 or 12: (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6). That is 6 of 36: 636≈0.167\frac{6}{36} \approx 0.167.

  • Answer: 0.1667

3. What is the probability that at least one of the dice shows a 6?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

The complement is "neither die shows a 6": 5×5=255 \times 5 = 25 outcomes. So P(at least one 6)P(\text{at least one } 6)=1−2536= 1 - \frac{25}{36}=1136= \frac{11}{36}≈0.306\approx 0.306. Counting directly: 6 in the red row, 6 in the blue column, but (6, 6) is in both, so 11.

  • Answer: 0.3056

Table

A club's members by class

A club has 40 members: 12 freshmen, 10 sophomores, 8 juniors and 10 seniors. One member is chosen at random to win a prize. (Invented data.)
Club members by class
ClassFreshmanSophomoreJuniorSeniorTotal
Members121081040

4. What is the probability that the winner is a junior?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Each of the 40 members is equally likely to be chosen, and 8 are juniors: 840=0.2\frac{8}{40} = 0.2.

  • Answer: 0.2

5. Which gives the sample space for the winner's class, with the correct probabilities?

  1. Freshman 0.25, sophomore 0.25, junior 0.25, senior 0.25
  2. Freshman 0.3, sophomore 0.25, junior 0.2, senior 0.25
  3. Freshman 12, sophomore 10, junior 8, senior 10
  4. Freshman 0.3, sophomore 0.25, junior 0.2, senior 0.2
Show the answer

The sample space lists every possible class. The probabilities come from the counts: 1240=0.3\frac{12}{40} = 0.3, 1040=0.25\frac{10}{40} = 0.25, 840=0.2\frac{8}{40} = 0.2, 1040=0.25\frac{10}{40} = 0.25; they add to 1.

  • Freshman 0.25, sophomore 0.25, junior 0.25, senior 0.25: Four classes does not make them equally likely; the counts differ.
  • Correct: Freshman 0.3, sophomore 0.25, junior 0.2, senior 0.25: Right: each class's count over 40.
  • Freshman 12, sophomore 10, junior 8, senior 10: Those are counts. Probabilities are shares between 0 and 1.
  • Freshman 0.3, sophomore 0.25, junior 0.2, senior 0.2: These add to 0.95, not 1. Seniors are 10 of 40, which is 0.25.

6. A fair coin is flipped 3 times. What is the probability of getting at least one head?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

The sample space has 8 equally likely outcomes (HHH, HHT, ..., TTT). Only TTT has no head, so P(at least one head)P(\text{at least one head})=1−18= 1 - \frac{1}{8}=0.875= 0.875.

  • Answer: 0.875

Part 9 · Summary

Summary

List the sample space. If its outcomes are equally likely, the probability of an event is its number of outcomes over the total. Probabilities run from 0 to 1 and the outcomes' probabilities add to 1. The complement rule, P(Ac)=1−P(A)P(A^c) = 1 - P(A), is often the fastest route, especially for "at least one".

Part 10 · Up next

What comes next

Part 11 · Connections

Connections