Algebra 2 Conditional Probability and Independence Lesson (Free PDF)

Algebra 2 math lesson: Conditional Probability and Independence – Maths PDF

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This probability lesson for Algebra 2 covers conditional probability and independence (Common Core S-CP.A.1, S-CP.A.2, S-CP.A.3, S-CP.A.4, S-CP.A.5, S-CP.B.6, S-CP.B.7, S-CP.B.8, S-CP.B.9), with a free PDF to print.

Conditional probability, P(A given B) = P(A and B) / P(B), is the probability of an event when another event is known to occur. Events are independent when P(A and B) = P(A)·P(B). Use the addition rule, the multiplication rule, two-way tables, tree diagrams, and permutations and combinations to find probabilities.

Algebra 2 · Common Core State Standards: S-CP.A.1, S-CP.A.2, S-CP.A.3, S-CP.A.4, S-CP.A.5, S-CP.B.6, S-CP.B.7, S-CP.B.8, S-CP.B.9

Key facts in this probability lesson

Key facts

P(not A) = 1 – P(A); P(A or B) = P(A) + P(B) – P(A and B).
P(A given B) = P(A and B) / P(B).
Independent: P(A and B) = P(A)·P(B).
Multiplication rule: P(A and B) = P(A)·P(B given A).
Permutations P(n, r) = n!/(n – r)!; combinations C(n, r) = n!/(r!(n – r)!).

How do you describe events and find their probabilities?

The sample space is the set of all possible outcomes, and an event is a subset of the sample space. For equally likely outcomes, \(P(A) = \frac{\text{number of outcomes in }A}{\text{total outcomes}}\). For two dice, there are 36 outcomes, and 6 of them have sum 7, so \(P(\text{sum is } 7) = \frac{6}{36} = \frac{1}{6}\).

All 36 outcomes for two dice; the sums of 7 are highlighted.

The complement of \(A\) is “not \(A\)”: \(P(\text{not } A) = 1 – P(A)\). The event “\(A\) or \(B\)” is the union and “\(A\) and \(B\)” is the intersection. The addition rule is \(P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)\). For example, among 52 cards, \(P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} – \frac{1}{52} = \frac{16}{52} = \frac{4}{13}\).

30 students: 18 in band (13 + 5), 12 in choir (5 + 7), 5 in both.

What is conditional probability?

Conditional probability and independence

The conditional probability of \(A\) given \(B\) is \(P(A | B) = \frac{P(A \text{ and } B)}{P(B)}\), the probability of \(A\) when you restrict attention to \(B\). Events are independent if knowing one does not change the other: \(P(A \text{ and } B) = P(A) \cdot P(B)\), equivalently \(P(A | B) = P(A)\).

Of 200 students, 80 play a sport. P(plays a sport | Grade 11) = 20/120.

In the table, \(P(\text{sport} | \text{Grade } 11) = \frac{20}{120} = \frac{1}{6}\), while \(P(\text{sport}) = \frac{50}{200} = \frac{1}{4}\). They are different, so the events are not independent. The multiplication rule is \(P(A \text{ and } B) = P(A) \cdot P(B | A)\).

Two draws from a bag of 5 red (R) and 3 blue (B) marbles, without replacement.

Drawing two marbles without replacement: \(P(\text{both red}) = \frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\). With replacement the draws are independent and the probability is \(\frac{5}{8} \cdot \frac{5}{8} = \frac{25}{64}\).

How do you count arrangements and selections?

When order matters, count permutations: \(P(n, r) = \frac{n!}{(n – r)!}\). Choosing 1st, 2nd, and 3rd place from 5 runners gives \(P(5, 3) = 5 \cdot 4 \cdot 3 = 60\). When order does not matter, count combinations: \(C(n, r) = \frac{n!}{r!\,(n – r)!}\). A team of 3 chosen from 8 people has \(C(8, 3) = 56\) possibilities. Use counting to find probabilities: the probability of one specific team is \(\frac{1}{56}\).

Probability lesson: questions and answers

What is conditional probability?

Conditional probability is the probability of an event given that another event has occurred. P(A given B) equals P(A and B) divided by P(B).

How do you tell if two events are independent?

Check whether P(A and B) equals P(A) times P(B). Equivalent: P(A given B) equals P(A).

What is the addition rule?

P(A or B) = P(A) + P(B) – P(A and B). Subtract the overlap so it is not counted twice.

What is the difference between permutations and combinations?

A permutation counts arrangements where order matters, such as first, second, third place. A combination counts selections where order does not matter, such as a committee.

How do you use a two-way table to find probabilities?

Divide the count in the cell by the total for a joint probability. For a conditional probability, divide by the total of the given row or column.

How to use this probability lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

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