Study Guide11th GradeLevel 2 · Standard

Free 11th Grade Probability: Tree Diagrams and Conditional Probability

A senior study guide to multi-stage probability: the addition and multiplication rules, tree diagrams with and without replacement, conditional probability and testing for independence. Worked examples include a two-draw bag problem, a two-way class survey, an independence check and a factory defect problem where the tree runs backwards, with a six-question self-check and answers.

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Grade: 11th Grade
Subject: Math
Topic: Probability
Difficulty: Level 2 · Standard
Estimated time: 25 minutes
Curriculum: Texas grade levels (TEKS)
Answers: Not applicable
Format: PDF (A4) + print

Students will practice

applying the addition rule and the complement to find probabilities
drawing tree diagrams for events with and without replacement and multiplying along branches
calculating conditional probabilities using P(A | B) = P(A ∩ B) ÷ P(B)
testing whether two events are independent

Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Probability: Tree Diagrams and Conditional Probability

Study guide

How to use this study guide

Read it together first, pausing at each worked example to try the step before reading the answer.
Attempt the "Check yourself" questions at the end without looking back.
Then practice with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

11th Grade students (typically ages 16–17) working on probability. It is pitched at level 2 · standard.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try Counting Techniques: Permutations and Combinations.

Success TutoringWorksheets11th Grade · Mathematics · Probability

Probability: Tree Diagrams and Conditional Probability

What you need to know

For equally likely outcomes, P(A) = [frac number of favorable outcomes/total number of outcomes], and the complement rule says P(not A) = P(A′) = 1 − P(A). The addition rule for "A or B" is P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the overlap is subtracted so it is not counted twice. If A and B cannot both happen (mutually exclusive), P(A ∩ B) = 0.

For two stages, draw a tree diagram: multiply along a path to find the probability of that sequence, and add across the paths that match the question. Events are independent when one does not change the probability of the other, and then P(A ∩ B) = P(A) × P(B). When one event does change the other (for example drawing without replacement), use conditional probability: P(B | A) means "the probability of B given that A has happened".

Rule	Formula	When to use it
Complement	P(A′) = 1 − P(A)	"at least one" questions: 1 − P(none)
Addition	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	A or B (or both)
Multiplication (general)	P(A ∩ B) = P(A) × P(B \| A)	A and then B — the second branch of a tree
Independence	P(A ∩ B) = P(A) × P(B)	only when A and B are independent (test it!)
Conditional	P(A \| B) =	"given that", "of those who…", "if it is known that…"

A bag holds 3 red and 2 blue marbles. Two are drawn without replacement.

First draw: P(R) = , P(B) = . Second draw has only 4 marbles left, and the colors depend on the first draw.
P(both red) = × = = [frac 3/10]
P(one of each) = P(RB) + P(BR) = × + × = + = = [frac 3/5]
P(at least one blue) = 1 − P(both red) = 1 − = [frac 7/10]
With replacement instead, every branch stays or , so P(both red) would be × = .

In a class of 30, 18 study Chemistry, 12 study Physics and 8 study both.

Chemistry only: 18 − 8 = 10. Physics only: 12 − 8 = 4. Neither: 30 − (10 + 8 + 4) = 8.
P(Chemistry or Physics) = = = [frac 11/15]
P(neither) = = [frac 4/15]
P(Physics | Chemistry) = = = [frac 4/9]. (Of the 18 Chemistry students, 8 also do Physics — the condition shrinks the total to 18.)

P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent?

Test: P(A) × P(B) = 0.4 × 0.5 = 0.2
This equals P(A ∩ B) = 0.2, so A and B are independent.
Equivalently P(A | B) = = 0.4 = P(A): knowing B happened does not change the chance of A.

Machine A makes 60% of a factory's parts and 2% of them are faulty. Machine B makes the other 40% and 5% are faulty.

Tree: first branch is the machine (0.6 / 0.4), second branch is faulty or not (0.02 / 0.98 after A, 0.05 / 0.95 after B).
P(faulty) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032
A faulty part is found. P(it came from A | faulty) = = = 0.375 (= )
Even though A makes most parts, less than half of the faulty parts come from A because its fault rate is lower.

Common mistake: forgetting that without replacement changes both the numerator and the denominator on the second draw. After one red is taken from 3 red and 2 blue, there are 2 red out of 4, not 3 out of 5.

Common mistake: using the whole sample space for a conditional probability. "Given that" means you only look at the outcomes where the condition is true, so the denominator becomes P(B), not 1. And "at least one" is almost always fastest through the complement: 1 − P(none).

Check yourself

P(A) = 0.3. Find P(A′).
A fair die is rolled twice. Find the probability of two sixes.
A bag holds 4 green and 6 yellow counters. Two are drawn without replacement. Find P(both green).
P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.1. Find P(A ∪ B).
Using the values in question 4, find P(A | B).
Using the values in question 4, are A and B independent? Explain.

Answers: 1. 0.7 2. × = 3. × = = 4. 0.5 + 0.4 − 0.1 = 0.8 5. = 0.25 6. No: P(A) × P(B) = 0.2, which is not equal to P(A ∩ B) = 0.1

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Created by: Success Tutoring
Last reviewed: 1 October 2026
How it was made: Written by our team

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