How to Solve Probability Problems? (+FREE Worksheet!)

Probability measures how likely an event is to occur, expressed as a number between 0 and 1 (or 0% and 100%). Understanding how to set up and solve probability problems is a key skill in Algebra 1 statistics. This guide covers the basic probability formula, simple and compound events, and includes worked examples, two video lessons, and practice problems so you can master the concept.

What Is Probability?

The probability of an event is the ratio of the number of favorable outcomes to the total number of equally likely outcomes. A probability of 0 means the event is impossible; a probability of 1 means the event is certain. All other probabilities fall strictly between 0 and 1.

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\(\color{blue}{P(\text{ event }) = (\text{ Number of favorable outcomes })}\) ÷ (Total number of possible outcomes)

Types of Probability Events

Simple Events

A simple event has a single outcome. Apply the basic formula directly.

Quick example: Rolling a standard 6-sided die, what is the probability of rolling a 4?
\(\color{blue}{P(4) = 1}\) ÷ 6

Compound Events — “And” (Intersection)

For independent events (the result of one does not affect the other), multiply the probabilities:

\(\color{blue}{P(A \text{ and B }) = P(A) \times P(B)}\)

Quick example: Flipping a fair coin twice: \(\color{blue}{P(\text{ Heads and Heads }) = (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4}}\)

Compound Events — “Or” (Union)

For mutually exclusive events (they cannot both happen at the same time), add the probabilities:

\(\color{blue}{P(A \text{ or B }) = P(A) + P(B)}\)

Quick example: Rolling a 2 or a 5 on a die: \(\color{blue}{P(2 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}}\)

Complementary Events

The complement of event A is everything that is NOT A. The two probabilities must sum to 1:

\(\color{blue}{P(\text{ not A }) = 1 – P(A)}\)

Quick example: \(\color{blue}{P(\text{ not rolling a } 3) = 1 – \frac{1}{6} = \frac{5}{6}}\)

Step-by-Step Summary

1. Identify the total number of equally likely outcomes in the sample space.
2. Count the number of favorable outcomes for the event.
3. Apply the formula: \(\color{blue}{P(\text{ event }) = \text{ favorable }}\) ÷ total.
4. For compound events: multiply probabilities for “and” (independent); add for “or” (mutually exclusive).
5. Use \(\color{blue}{P(\text{ not A }) = 1 – P(A)}\) for complementary events.
6. Simplify the fraction and, if needed, convert to a decimal or percentage.

Watch: Basic Probability (Video Lesson)

Math Antics introduces probability with clear, visual explanations:

Probability Problems – Worked Examples

Example 1: A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. A marble is drawn at random. Find the probability of drawing a red marble.

Total \(\color{blue}{\text{ outcomes } = 5 + 3 + 2 = 10}\)
Favorable outcomes \(\color{blue}{(\text{ red }) = 5}\)
\(\color{blue}{P(\text{ red }) = 5}\) ÷ 10 = \(\color{blue}{\frac{1}{2}}\)

Example 2: Using the same bag, find the probability of NOT drawing a blue marble.

\(\color{blue}{P(\text{ blue }) = 3}\) ÷ 10
\(\color{blue}{P(\text{ not blue }) = 1 – \frac{3}{10}}\) = \(\color{blue}{\frac{7}{10}}\)

Example 3: A fair coin is flipped twice. What is the probability of getting exactly one head?

Sample space: HH, HT, TH, TT → 4 equally likely outcomes
Favorable outcomes (exactly one head): HT, TH → 2 outcomes
\(\color{blue}{P(\text{ exactly one head }) = 2}\) ÷ 4 = \(\color{blue}{\frac{1}{2}}\)

Example 4: A 6-sided die is rolled. What is the probability of rolling an even number or a number greater than 4?

Even numbers: {2, 4, 6}; Numbers > 4: {5, 6}. Union: {2, 4, 5, 6} → 4 outcomes
\(\color{blue}{P(\text{ even or } > 4) = 4}\) ÷ 6 = \(\color{blue}{\frac{2}{3}}\)

More Practice: Independent and Dependent Events (Video Lesson)

Khan Academy explains independent and dependent events with detailed examples:

Exercises: Probability Problems

1. A bag holds 4 yellow, 6 orange, and 10 purple chips. What is P(yellow)?
2. A spinner has 8 equal sections numbered 1–8. What is P(prime number)?
3. A fair coin is flipped and a die is rolled. What is P(heads and 6)?
4. A deck of 52 cards is shuffled. What is P(drawing a king or a queen)?
5. The probability of rain on Saturday is 0.3. What is the probability it does NOT rain?
6. Two dice are rolled. What is P(\(\color{blue}{\text{ sum } = 7}\))?

Answers

1. 4 ÷ 20 = \(\color{blue}{\frac{1}{5}}\)
2. Prime numbers 1–8: {2, 3, 5, 7} → 4 ÷ 8 = \(\color{blue}{\frac{1}{2}}\)
3. \(\color{blue}{P(\text{ heads }) \times P(6) = \frac{1}{2} \times \frac{1}{6}}\) = \(\color{blue}{\frac{1}{12}}\)
4. \(\color{blue}{P(\text{ king }) + P(\text{ queen }) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52}}\) = \(\color{blue}{\frac{2}{13}}\)
5. \(\color{blue}{1 – 0.3}\) = 0.7
6. Pairs summing to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 ÷ 36 = \(\color{blue}{\frac{1}{6}}\)

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Frequently Asked Questions

How do you solve probability problems step by step?

First list or count every possible outcome, then count the outcomes you want, and write probability = favorable ÷ total. Simplify and check that the answer is between 0 and 1. For example, rolling a number greater than 4 on a fair die has 2 favorable outcomes (5 and 6) out of 6, so the probability is 2/6 = 1/3.

How do you solve dice probability problems?

For one fair die there are 6 equally likely outcomes; for two dice there are 6 × 6 = 36. Count the pairs that fit the event. A sum of 7 comes from (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), so P(sum of 7) = 6/36 = 1/6. Making a 6-by-6 table of sums keeps the counting organized.

How do you solve card probability problems?

Use the structure of a standard deck: 52 cards, 4 suits of 13, and 4 of each rank. P(ace) = 4/52 = 1/13 and P(heart) = 13/52 = 1/4. If you draw two cards without replacement, the second draw changes: P(two aces) = 4/52 × 3/51 = 1/221. Always check whether cards are replaced before multiplying.

How do you solve coin probability problems?

Each flip of a fair coin has P(heads) = 1/2, and flips are independent, so multiply for several flips. Three heads in a row has probability 1/2 × 1/2 × 1/2 = 1/8. For at least one head, use the complement: 1 − P(no heads) = 1 − 1/8 = 7/8. Listing outcomes like HHT and HTH helps for small cases.

How do you solve probability problems using a tree diagram?

Draw a branch for each possible outcome of the first event, then branches for the second event from each end. Write the probability on every branch, multiply along a path, and add the paths you need. Drawing two marbles from a bag of 2 red and 3 blue without replacement, P(red then blue) = 2/5 × 3/4 = 6/20 = 3/10.

How do you solve conditional probability problems?

Conditional probability asks for the chance of A given that B already happened: P(A | B) = P(A and B) ÷ P(B). In practice, shrink the sample space to B. If 12 students in a class play a sport and 8 of those also play an instrument, then P(instrument | sport) = 8/12 = 2/3.

Related Topics

• Permutations and Combinations
• Mean, Median, Mode, and Range
• Pie Graph
• Scatter Plots
• Statistics