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# 12.11: Counting Rule to Calculate Probabilities

Difficulty Level: At Grade Created by: CK-12
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Practice Counting Rule to Calculate Probabilities

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Rebecca loves green Skittles more than all the other colors:  red, yellow, orange, and purple.  She wonders if she places a Skittle of each color in a bowl, five Skittles total, and pulls one Skittle, replaces it, then pulls one again, what are her chances of pulling a green Skittle each time.

In this concept, you will learn how to calculate the probability of two separate events producing favorable outcomes.

### Using the Counting Principle to Calculate Probabilities

You can use the Counting Principle to find the number of total outcomes for two separate events, for example, two spins with two spinners. Considering the spinners below, what is the probability that you spin each spinner and they both land on the same color?

There are four outcomes on one spinner and three outcomes on the other spinner.

Total outcomes=4 outcomes3 outcomes=12 outcomes\begin{align*}\text{Total outcomes} &= 4 \ \text{outcomes} \cdot 3 \ \text{outcomes}\\ &= 12 \ \text{outcomes}\end{align*}

Now list those 12 outcomes and mark the outcomes that are the same color for both spins.

red-redred-bluered-greenblue-redblue-blueblue-greenyellow-redyellow-blueyellow-greengreen-redgreen-bluegreen-green\begin{align*}& \text{red-red} && \text{blue-red} && \text{yellow-red} && \text{green-red}\\ & \text{red-blue} && \text{blue-blue} && \text{yellow-blue} && \text{green-blue}\\ & \text{red-green} && \text{blue-green} && \text{yellow-green} && \text{green-green}\end{align*}

Since there are 3 outcomes that have the same color for both spins:

P(same)=312=14\begin{align*}P (\text{same}) = \frac{3}{12}=\frac{1}{4}\end{align*}

The probability of both spinners landing on the same color is 14\begin{align*}\frac{1}{4}\end{align*}.

Let's look at another example.

Anna flips a coin 3 times in a row. What is the probability that she will get heads all 3 times?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:

Total outcomes=2 outcomes2 outcomes2 outcomes=8 outcomes\begin{align*}\text{Total outcomes} &= 2 \ \text{outcomes} \cdot 2 \ \text{outcomes} \cdot 2 \ \text{outcomes}\\ &= 8 \ \text{outcomes}\end{align*}

Next, list all 8 outcomes and find the number of ways Anna can get heads all 3 times. There is only one arrangement in which all 3 flips result in heads.

heads-heads-headsheads-heads-tailsheads-tails-headsheads-tails-tailstails-heads-headstails-heads-tailstails-tails-headstails-tails-tails\begin{align*}& \text{heads-heads-heads} && \text{tails-heads-heads}\\ & \text{heads-heads-tails} && \text{tails-heads-tails}\\ & \text{heads-tails-heads} && \text{tails-tails-heads}\\ & \text{heads-tails-tails} && \text{tails-tails-tails}\end{align*}

Then, find the ratio of favorable outcomes to total outcomes:

P(red-red-red)=18\begin{align*}P (\text{red-red-red}) = \frac{1}{8}\end{align*}

The answer is the probability of Anna flipping heads all three times is 18\begin{align*}\frac{1}{8}\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about Rebecca's curiosity about pulling out her favorite color Skittle.

She will replace the Skittle after the first pull before pulling the second time.  The colors in the bowl are red, orange, yellow, green, and purple.  What is the probability that Anna will pull the green Skittle both times?

First, find the number of total outcomes by multiplying the number of outcomes for each pull:Total outcomes=5 outcomes5 outcomes=25 outcomes\begin{align*}\text{Total outcomes} &= 5 \ \text{outcomes} \cdot 5 \ \text{outcomes}\\ &= 25 \ \text{outcomes}\end{align*}Next, list all 25 outcomes and find the number of ways Rebecca can pull the green Skittle both times. There is only one arrangement in which boht pulls result in green Skittles.

red-red                  red-orange                red-yellow            red-green

red-purple

orange-orange       orange-red                orange-yellow       orange-green

orange-purple

yellow-yellow         yellow-orange            yellow-red            yellow-green

yellow-purple

green-green           green-yellow             green-orange        green-red

green-purple

purple-purple         purple-green              purple-yellow        purple-orange

purple-red

Then, find the ratio of favorable outcomes to total outcomes:

P(green-green)=125\begin{align*}P (\text{green-green}) = \frac{1}{25}\end{align*}

The answer is the probability of pulling a green Skittle both pulls is 125\begin{align*}\frac{1}{25}\end{align*}.

#### Example 2

If Sam flips a coin five times. What is the probability of having it come up heads all at the same time?

First, figure out the total number of outcomes:

There can be two possible outcomes each time Sam flips the coin.

Total outcomes = 2 outcomes x 2 outcomes x 2 outcomes x 2 outcomes x 2 outcomes = 32 outcomes

There are 32 possible outcomes.

Next, list all of the outcomes:

H-H-H-H-H         H-H-H-H-T            H-H-H-T-H           H-H-H-T-T            H-H-T-H-H

H-H-T-H-T          H-H-T-T-H            H-H-T-T-T           H-T-H-H-H            H-T-H-H-T

H-T-H-T-H          H-T-H-T-T            H-T-T-H-H           H-T-T-H-T             H-T-T-T-H

H-T-T-T-T

T-H-H-H-H         T-H-H-H-T            T-H-H-T-H            T-H-H-T-T            T-H-T-H-H

T-H-T-H-T         T-H-T-T-H            T-H-T-T-T             T-T-H-H-H            T-T-H-H-T

T-T-H-T-H         T-T-H-T-T             T-T-T-H-H            T-T-T-H-T             T-T-T-T-H

T-T-T-T-T

Then, determine how many times all heads appears and create a ratio of favorable outcomes to total outcomes:

H-H-H-H-H only appears one time out of 32 total outcomes = 132\begin{align*}\frac{1}{32}\end{align*}.

The answer is the probability of Sam flipping all heads over five flips is 132\begin{align*}\frac{1}{32}\end{align*}.

#### Example 3

Abra flips a coin 2 times. What is the probability that both flips will match?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:

Total outcomes=2 outcomes2 outcomes=4 outcomes\begin{align*}\text{Total outcomes} &= 2 \ \text{outcomes} \cdot 2 \ \text{outcomes}\\ &= 4 \ \text{outcomes}\end{align*}Next, list all 4 outcomes and find the number of times Abra can get matching flips . There are two outcomes in which the flips will match:heads-headsheads-tailstails-headstails-tails\begin{align*}& \text{heads-heads} && \text{tails-heads}\\ & \text{heads-tails} && \text{tails-tails}\\ \end{align*}Then, find the ratio of favorable outcomes to total outcomes: P(heads-heads or tails-tails)=24=12\begin{align*}P (\text{heads-heads or tails-tails}) = \frac{2}{4} = \frac{1}{2}\end{align*}

The answer is the probability of Abra having matching flips is \begin{align*}\frac{1}{2}\end{align*}.

#### Example 4

Abra flips a coin 2 times. What is the probability that both flips will NOT match?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:\begin{align*}\text{Total outcomes} &= 2 \ \text{outcomes} \cdot 2 \ \text{outcomes}\\ &= 4 \ \text{outcomes}\end{align*}Next, list all 4 outcomes and find the number of times the outcomes will not result in matching flips . There are two outcomes in which the flips will not match:\begin{align*}& \text{heads-heads} && \text{tails-heads}\\ & \text{heads-tails} && \text{tails-tails}\\ \end{align*}Then, find the ratio of favorable outcomes to total outcomes: \begin{align*}P (\text{heads-tails or tails-heads}) = \frac{2}{4} = \frac{1}{2}\end{align*}

The answer is the probability of Abra not having matching flips is \begin{align*}\frac{1}{2}\end{align*}.

#### Example 5

Abra flips a coin 3 times. What is the probability that all 3 flips will match?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:

\begin{align*}\text{Total outcomes} &= 2 \ \text{outcomes} \cdot 2 \ \text{outcomes} \cdot 2 \ \text{outcomes}\\ &= 8 \ \text{outcomes}\end{align*}Next, list all 8 outcomes and find the number of ways Abra can get matching flips all 3 times. There are two arrangements in which all 3 flips result in a match.

\begin{align*}& \text{heads-heads-heads} && \text{tails-heads-heads}\\ & \text{heads-heads-tails} && \text{tails-heads-tails}\\ & \text{heads-tails-heads} && \text{tails-tails-heads}\\ & \text{heads-tails-tails} && \text{tails-tails-tails}\end{align*}Then, find the ratio of favorable outcomes to total outcomes:

\begin{align*}P (\text{heads-heads-heads or tails-tails-tails}) = \frac{2}{8} = \frac{1}{4}\end{align*}

The answer is the probability of Anna flipping heads all three times is \begin{align*}\frac{1}{4}\end{align*}.

### Review

Find probabilities using the Counting Principle.

1. Abra flips a coin 3 times. What is the probability that all 3 flips will NOT match?
2. Abra flips a coin 3 times. What is the probability that heads will come up exactly 2 times?

1. Billy spins the spinner twice. What is the probability that blue will come up both times?
2. Billy spins the spinner twice. What is the probability that blue will come up at least one time?
3. Billy spins the spinner twice. What is the probability that blue will come up exactly one time?
4. Billy spins the spinner twice. What is the probability that both spins will match?
5. Cindy tosses a number cube two times. What is the probability that both tosses will match?
6. Cindy tosses a number cube two times. What is the probability that both tosses will NOT match?
7. Cindy tosses a number cube two times. What is the probability that the sum of the two tosses will be 7?
8. Cindy tosses a number cube two times. What is the probability that the number will be three?
9. Tyler rolls a number cube three times. What is the probability that he will roll a two?
10. Tyler rolls a number cube three times. What is the probability that he will roll an odd number?
11. What is the probability that he will roll an even number?
12. If Tyler rolls a number cube once, what is the probability of rolling a two or a five?
13. If Tyler rolls a number cube six times, what is the probability of rolling a two or a five?

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