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Dependent Events

Two outcomes both occurring dependently.

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Dependent Events

Have you ever tried magic tricks?

Casey has been reading a book on magic. He is especially interested in card tricks. Casey has a pile of cards. There are four Jacks, four Queens and four Kings. What is the probability that Casey will pull two queens out of the pile?

Do you know how to do this?

This Concept is about probability and about calculating the probability of two dependent events both occurring. You will learn what you need to know in this Concept.

Guidance

Previously we worked on how to find the probability of two independent events. Well, we can use a similar method to find the probability of two dependent events.

If we go back to our laundry bag full of socks and ask: What is the probability of picking two socks from the bag and have them both turn out to be 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*}

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*}

This gives a probability of both events occurring as:

P(red 1st sock and red 2nd sock)=P(red 1st sock)P(red 2nd sock)=1225=210=15

The same general method works for calculating any two (or more) dependent events.

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.

The probability of the 1st Jack is:

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

Once the 1st Jack is taken, the probability of 2nd Jack is only 4 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*}

So:

P(1st Jack and 2nd Jack)=P(1st Jack)P(2nd Jack)=5847=514

The answer is 514\begin{align*}\frac{5}{14}\end{align*}.

We can use the Probability Rule to find the probability of both independent and dependent events!

Try a few of these on your own.

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

Example A

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

Solution: 115\begin{align*}\frac{1}{15}\end{align*}

Example B

What is the probability of pulling out two blue marbles?

Solution: 215\begin{align*}\frac{2}{15}\end{align*}

Example C

What is the probability of pulling out two yellow marbles?

Solution: 115\begin{align*}\frac{1}{15}\end{align*}

Here is the original problem once again.

Casey has been reading a book on magic. He is especially interested in card tricks. Casey has a pile of cards. There are four Jacks, four Queens and four Kings. What is the probability that Casey will pull two queens out of the pile?

To figure this out let's begin by writing a ratio to show the number of queens compared to the total number of cards.

412\begin{align*}\frac{4}{12}\end{align*}

This is the chance of pulling the first queen.

Next, we can write the probability of the second queen.

311\begin{align*}\frac{3}{11}\end{align*}

Now we can multiply them.

312×311=9132\begin{align*}\frac{3}{12} \times \frac{3}{11} = \frac{9}{132}\end{align*}

We can simplify this fraction.

344\begin{align*}\frac{3}{44}\end{align*}

Finally, we can change this to a percent.

.068=6.8%\begin{align*}.068 = 6.8\%\end{align*}

Guided Practice

Here is one for you to try on your own.

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?

To figure this out, we first have to figure out the total number of students.

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

Next, we write a ratio to show the chances of her picking the first boy.

813\begin{align*}\frac{8}{13}\end{align*}

Next, we write the ratio of choosing the second boy.

712\begin{align*}\frac{7}{12}\end{align*}

Now we multiply.

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

1439\begin{align*}\frac{14}{39}\end{align*}

Explore More

Directions: 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?

Directions: 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?

Vocabulary Language: English

Dependent Events

Dependent Events

In probability situations, dependent events are events where one outcome impacts the probability of the other.
Independent Events

Independent Events

Two events are independent if the occurrence of one event does not impact the probability of the other event.
Probability Rule

Probability Rule

The probability of two independent events A and B both occurring is P( A and B) = P(A) $\cdot$ P(B).