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# Dependent Events

## Two outcomes both occurring dependently.

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Dependent Events

Antonio is responsible for washing the dishes.  There are four bowls, three plates, and six cups in the sink.  If Antonio washes a bowl first, what is the probability that he will grab another bowl next?

In this concept, you will learn how to calculate the probability of two dependent events occurring.

### Guidance

The Probability Rule is the probability that two independent eventsA\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, will both occur and is written as the formula:P(A and B)=P(A)P(B)\begin{align*}P(A \ \text{and} \ B) = P (A) \cdot P (B)\end{align*}

If you pull two socks from a bag full of the following blue and red socks, what is the probability of both socks being red?

The probability of the first sock being red is:

P(red 1st sock)=36=12\begin{align*}P (\text{red 1st sock}) = \frac{3}{6} = \frac{1}{2}\end{align*}

What about the second sock?  Having removed the first sock from the bag, we have changed the number of red socks and the total number of socks in the bag. So now instead of there being 3 red socks out of 6 total socks, there are only 2 red socks left out of 5 total socks:

P(red 2nd sock)=25\begin{align*}P (\text{red 2nd sock}) = \frac{2}{5}\end{align*}

The probability of two red socks being pulled is:

P(red 1st sock and red 2nd sock)=P(red 1st sock)P(red 2nd sock)=1225=210=15\begin{align*}P (\text{red 1st sock and red 2nd sock}) & = P (\text{red 1st sock}) \cdot P (\text{red 2nd sock})\\ & = \frac{1}{2} \cdot \frac{2}{5}\\ & = \frac{2}{10}\\ & = \frac{1}{5}\end{align*}

Let's consider another example.

A stack of 8 cards has 4 Jacks and 4 Queens. What is the probability of picking 2 Jacks from the stack at random?

Use the Probability Rule to find the probability of the two dependent events.

First, determine the probability of the 1st Jack:

P(1st Jack)=58\begin{align*}P (\text{1st Jack}) = \frac{5}{8}\end{align*}

Next, determine the probability of the 2nd Jack:

Once the 1st Jack is taken, the probability of 2nd Jack is only 3 of 7 because there are only 4 Jacks left out of 7 total cards:

P(2nd Jack)=47\begin{align*}P (\text{2nd Jack}) = \frac{4}{7}\end{align*}

Then, substitute the values into the Probability Rule formula:

P(1st Jack and 2nd Jack)=P(1st Jack)P(2nd Jack)=5847=514\begin{align*}P (\text{1st Jack and 2nd Jack}) & = P (\text{1st Jack}) \cdot P (\text{2nd Jack})\\ & = \frac{5}{8} \cdot \frac{4}{7}\\ & = \frac{5}{14}\end{align*}

The answer is the probability of picking two Jacks is 514\begin{align*}\frac{5}{14}\end{align*}.

### Guided Practice

There are five girls and eight boys in a group. Mrs. Marsh is going to choose two students randomly to lead the line. What is the probability that she will choose two boys?

First, figure out the total number of students:

5+8=13\begin{align*}5 + 8 = 13\end{align*}

Next, write a ratio to show the probability of her picking the first boy and then the second boy:

probability of picking first boy = 813\begin{align*}\frac{8}{13}\end{align*}

probability of picking second boy = 712\begin{align*}\frac{7}{12}\end{align*}

Then, multiply the two values together:

813×712=1439\begin{align*}\frac{8}{13} \times \frac{7}{12}=\frac{14}{39}\end{align*}

The answer is the probability of Mrs. Marsh choosing two boys is 1439\begin{align*}\frac{14}{39}\end{align*}.

### Examples

A bag has three red marbles, two yellow marbles and four blue marbles.

#### Example 1

What is the probability of pulling two red marbles out of the bag?

First, figure out the total number of marbles:3+2+4=9\begin{align*}3 + 2 + 4 = 9\end{align*}

Next, write a ratio to show the probability of pulling the first red marble and the second red marble out of the bag:

the ratio of the probability of pulling the first red marble: 39=13\begin{align*}\frac{3}{9}= \frac{1}{3}\end{align*}

the ratio of the probability of pulling the second red marble:  28=14\begin{align*}\frac{2}{8} = \frac{1}{4}\end{align*}

Then, multiply the two values together:

13×14\begin{align*}\frac{1}{3} \times \frac{1}{4}\end{align*} = 112\begin{align*}\frac{1}{12}\end{align*}

The answer is the probability of choosing two red marbles is 112\begin{align*}\frac{1}{12}\end{align*}.

#### Example 2

What is the probability of pulling out two blue marbles?

First, figure out the total number of marbles:3+2+4=9\begin{align*}3 + 2 + 4 = 9\end{align*}

Next, write a ratio to show the probability of pulling the first blue marble and the second blue marble out of the bag:

the ratio of the probability of pulling the first blue marble:  49\begin{align*}\frac{4}{9}\end{align*}

the ratio of the probability of pulling the second blue marble:  38\begin{align*}\frac{3}{8}\end{align*}

Then, multiply the two values together: 49x38=1272=16\begin{align*}\frac{4}{9} x \frac{3}{8} = \frac{12}{72}= \frac{1}{6}\end{align*}

The answer is the probability of choosing two blue marbles is  16\begin{align*}\frac{1}{6}\end{align*}.

#### Example 3

What is the probability of pulling out two yellow marbles?

First, figure out the total number of marbles:3+2+4=9\begin{align*}3 + 2 + 4 = 9\end{align*}

Next, write a ratio to show the probability of pulling the first yellow marble and the second yellow marble out of the bag:

the ratio of the probability of pulling the first yellow marble: 29\begin{align*}\frac{2}{9}\end{align*}

the ratio of the probability of pulling the second yellow marble:  18\begin{align*}\frac{1}{8} \end{align*}

Then, multiply the two values together:29×18\begin{align*}\frac{2}{9} \times \frac{1}{8}\end{align*} = 272=136\begin{align*}\frac{2}{72}= \frac{1}{36}\end{align*}

The answer is the probability of choosing two yellow marbles is 136\begin{align*}\frac{1}{36}\end{align*}.

Remember Antonio washing the dishes?  What is the probability that he will wash a bowl first and then grab another bowl next when there are four bowls, three plates, and six cups in the sink?

First, figure out the total number of dishes in the sink: 4+3+6=13\begin{align*}4 + 3 + 6 = 13\end{align*}

Next, write a ratio to show the probability of picking the first bowl and then the second bowl:

probability of picking first bowl = 413\begin{align*}\frac{4}{13}\end{align*}

probability of picking second bowl = 312=14\begin{align*}\frac{3}{12} = \frac{1}{4}\end{align*}

Then, multiply the two values together:

413×14\begin{align*}\frac{4}{13} \times \frac{1}{4}\end{align*} = 452=113\begin{align*}\frac{4}{52}=\frac{1}{13}\end{align*}

The answer is the probability of Antonio choosing two bowls is  113\begin{align*}\frac{1}{13}\end{align*}.

### Explore More

Use this description to figure out the probability of each dependent event.

A box has eight kittens in it. Three calico, two white and three black.

1. What is the probability of choosing two white kittens?

2. What is the probability of choosing three black kittens?

3. What is the probability of choosing two black kittens?

4. What is the probability of choosing two calico kittens?

5. What is the probability of choosing three calico kittens?

6. What is the probability of choosing one white and then one black kitten?

7. What is the probability of choosing two calico and one black?

8. What is the probability of choosing one calico and one white kitten?

9. What is the probability of choosing a striped kitten?

10. What is the probability of choosing one white kitten and two black kittens?

Solve each problem.

11. A clothes dryer contains 5 black socks and 1 white sock. What is the probability of taking two socks, one after another, out of the dryer and having them both be black?

12. A clothes dryer contains 4 black socks and 2 white socks. What is the probability of taking two socks out of the dryer and having them both be black?

13. A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having them both be black?

14. A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having the first one be black and the second one be white?

15. Bob bought two theater box tickets. The computer randomly assigns the tickets in one of 5 seats: end seat A, middle seat B, middle seat C, middle seat D, or end seat E. What is the probability that the first ticket is A and the second ticket is seat B?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.18.

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### Vocabulary Language: English

TermDefinition
Dependent Events In probability situations, dependent events are events where one outcome impacts the probability of the other.
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
Probability Rule The probability of two independent events A and B both occurring is P( A and B) = P(A) $\cdot$ P(B).