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5. Probability
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Grade 8

5.02 Probability of independent events

Lesson
Practice

Probability of independent events

Two events are independent if the outcome of each event does not affect the outcome of the other event. For example, when we roll a die, there is always a \dfrac{1}{6} chance that it will land on 3. The outcome of the first roll will have no effect on outcome of the second roll.

The probability of a single event is the ratio of the number of favorable outcomes to the total number of outcomes:\text{Probability} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

The probability of two independent events is given by the formula:P\left(A\text{ and } B\right)=P\left(A\right) \cdot P\left(B\right)This means that the probability of both A and B occurring can be found by multiplying the probabilities of each event occurring separately.

Tree diagrams can be helpful for calculating the probabilities to two events. We have seen that tree diagrams can be used to list the sample space of an experiment. However, it can be difficult to list the outcomes in a simple tree diagram when there are many outcomes.

Exploration

Consider the tree diagrams shown.

Tree diagram without probabilities on the branches. See your teacher for more info.

Without probabilities on the branches

Tree diagram with probabilities on the branches. See your teacher for more info.

With probabilities on the branches

  1. Could these tree diagrams represent the same situation? Explain.

  2. For the tree diagram without probabilities on the branches, how could you find the probability of CC?

  3. Knowing the probability of CC from question 2, how could you find the probability of CC from the other tree diagram?

  4. What are the benefits and drawbacks of each of the diagrams?

In a probability tree diagram, the probability of each outcome is written on the branches.

A tree diagram showing 3 outcomes with different probabilities: 60% for W, 30% for L and 10% for D.

In this example, there is only one event with 3 possible outcomes. Each outcome has a different likelihood of occurring.

If you add the probabilities in each trial, the sum of the branches equals to 1 (or 100\%). This helps us check that all the outcomes are listed.

Here is an example of an experiment with 2 events.

A tree diagram with 0.7 probability for D and 0.3 probability for N. Both the first D and N branches have D and N branches with the same probability

This probability tree diagram shows the probability that a child gets home during the day \left(D\right) or at night \left(N\right) over two different days.

For each day, the probability that a child gets home during the day is P\left(D\right)=0.7, and the probability that a child gets home during the night is P\left(N\right)=0.3.

When using a probability tree diagram to find the probability of an outcome that includes both events, we must multiply across the branches of the tree diagram. For example, the probability that a child returns home during the day for both days is\begin{aligned}P\left(D\text{ and }D\right)&=P\left(D\right)\cdot P\left(D\right)\\&=0.7 \cdot 0.7 \\&= 0.49\end{aligned}The probability a child returns home at night on the first day and during the day on the second day is \begin{aligned}P\left(N\text{ and }D\right)&=P\left(N\right)\cdot P\left(D\right)\\&=0.3 \cdot 0.7\\&= 0.21\end{aligned}

When we add the probabilities of all the possible outcomes, it should be equal to 1 (or 100\%).

Examples

Example 1

A standard six-sided die is rolled 691 times.

If it lands on a six 108 times, what is the probability that the next roll will land on a six?

Worked Solution
Create a strategy

To find the probability that the next roll will land on a six, we need to determine if the next roll is dependent on the previous rolls. This will help us determine whether we need to find the theoretical probability or the experimental probability.

Apply the idea

The previous results do not affect the outcome of the next roll, so we will find the theoretical probability of rolling a 6.

\displaystyle P\left(A\right)\displaystyle =\displaystyle \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}Theoretical probability formula
\displaystyle =\displaystyle \dfrac{1}{6}Substitute the number of outcomes

Example 2

Two events A and B are such that:

  • P\left(A\right)=0.5
  • P\left(B\right)=0.7
  • P\left(A\text{ and } B\right)=0.3

Determine whether events A and B are independent.

Worked Solution
Create a strategy

Use the fact that if the events A and B are independent, then P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\right).

Apply the idea
\displaystyle P\left(A \text{ and } B\right)\displaystyle =\displaystyle P\left(A\right)\cdot P\left(B\right)
\displaystyle 0.3\displaystyle =\displaystyle 0.5 \cdot 0.7Substitute the values
\displaystyle 0.3\displaystyle \neq\displaystyle 0.35Evaluate

This shows that events A and B are not independent events.

Example 3

A standard deck of 52 cards has four queens. Ilham shuffles the deck then selects a card. He puts the card back in the deck, then selects a second card.

Let Q represent the event of drawing a queen and N represent the event of not drawing a queen. The tree diagram shows all the possible outcomes and probabilities:

A probability tree with multiple branches. Ask your teacher for more information.
a

Find the probability that both cards are queens.

Worked Solution
Create a strategy

Because the first card is replaced before drawing the second card, we know the events of drawing two cards are independent. The probability that both cards are queens can be represented by P\left(Q\text{ and }Q\right)=P\left(Q\right)\cdot P\left(Q\right).

Apply the idea

To find P\left(Q\text{ and }Q\right) using the tree diagram, we will multiply the probabilities on the branches with Q's. The tree shows that the probability of drawing a queen on each draw is \dfrac{4}{52} because there are 52 cards in the deck and 4 of them are queens.

\displaystyle P\left(Q\text{ and }Q\right)\displaystyle =\displaystyle P\left(Q\right)\cdot P\left(Q\right)Probability of drawing two queens
\displaystyle =\displaystyle \dfrac{4}{52}\cdot \dfrac{4}{52}Substitute known values
\displaystyle =\displaystyle \dfrac{1}{13}\cdot \dfrac{1}{13}Reduce the fractions
\displaystyle =\displaystyle \dfrac{1}{169}Multiply
Reflect and check

The probability of drawing two queens is about 0.6\% which is very unlikely.

b

Find the probability that neither card is a queen.

Worked Solution
Create a strategy

The probability that neither card is a queen can be represented by P\left(N\text{ and }N\right)=P\left(N\right)\cdot P\left(N\right). To find this probability using the tree diagram, we will multiply the probabilities on the branches with N's.

Apply the idea

The tree shows that the probability of not drawing a queen on each draw is \dfrac{48}{52} because there are 52 cards in the deck and 52-4=48 of them are not queens.

\displaystyle P\left(N\text{ and }N\right)\displaystyle =\displaystyle P\left(N\right)\cdot P\left(N\right)Probability of drawing two queens
\displaystyle =\displaystyle \dfrac{48}{52}\cdot \dfrac{48}{52}Substitute known values
\displaystyle =\displaystyle \dfrac{12}{13}\cdot \dfrac{12}{13}Reduce the fractions
\displaystyle =\displaystyle \dfrac{144}{169}Multiply
Reflect and check

Reducing the fractions before multiplying helps us work with smaller values, which can also help us avoid mistakes. However, we could have multiplied first, then reduced the result. We should arrive at the same answer either way:

\displaystyle P\left(N\text{ and }N\right)\displaystyle =\displaystyle P\left(N\right)\cdot P\left(N\right)Probability of drawing two queens
\displaystyle =\displaystyle \dfrac{48}{52}\cdot \dfrac{48}{52}Substitute known values
\displaystyle =\displaystyle \dfrac{2304}{2704}Multiply
\displaystyle =\displaystyle \dfrac{144}{169}Reduce by 16
c

Find the probability that Ilham selects only one queen.

Worked Solution
Create a strategy

The order in which the queen is drawn has not been specified, so there are two ways that Ilham can select only one queen:

  • Drawing the queen first, then not drawing a queen \left(Q\text{ and }N\right)

  • Not drawing a queen first, then drawing a queen \left(N\text{ and }Q\right)

We will calculate the probability of each situation separately, then add the results.

Apply the idea
\displaystyle P\left(Q\text{ and }N\right)+P\left(N\text{ and }Q\right)\displaystyle =\displaystyle P\left(Q\right)\cdot P\left(N\right) + P\left(N\right)\cdot P\left(Q\right)Probability of only one queen
\displaystyle =\displaystyle \left(\frac{4}{52}\cdot \frac{48}{52}\right)+\left(\frac{48}{52}\cdot \frac{4}{52}\right)Substitute known values
\displaystyle =\displaystyle \left(\frac{1}{13}\cdot \frac{12}{13}\right)+\left(\frac{12}{13}\cdot \frac{1}{13}\right)Simplify the fractions
\displaystyle =\displaystyle \frac{12}{169}+\frac{12}{169}Evaluate the multiplication
\displaystyle =\displaystyle \frac{24}{169}Evaluate the sum
Reflect and check

The probabilities we found in parts (a), (b), and (c) represent all the possible outcomes in the sample space. This means their probabilies will sum to 1.\dfrac{1}{169}+\dfrac{24}{169}+\dfrac{144}{169}=1

Idea summary

Two events are independent if the outcome of the each event does not affect the outcome of the other event.

If two events are independent, the probability of both events occurring is: P\left(A \text{ and } B\right) = P\left(A\right) \cdot P\left(B\right)This formula can also be used to check if two events are independent.

Outcomes

8.PS.1

The student will use statistical investigation to determine the probability of independent and dependent events, including those in context.

8.PS.1a

Determine whether two events are independent or dependent and explain how replacement impacts the probability.

8.PS.1c

Determine the probability of two independent events.

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