If you have six blue scarves in your closet and two red scarves, what is the probability that you'll pick a red scarf if you choose one without looking? Once you've taken the first scarf out of your closet and you pick a second one without looking what is the probability that you'll pick a blue scarf? Are these events independent or dependent?

### Watch This

Khan Academy Probability of Dependent Events

### Guidance

For 2 events to be dependent, the probability of the second event is dependent on the probability of the first event. In English, remember, the term dependent means to be unable to do without. This is similar to the mathematical definition of **dependent events**, where the probability of the second event occurring is influenced by the first event occurring.

#### Example A

Remember in the previous concept when you were asked to determine the probability of drawing 2 sevens from a standard deck of cards? What would happen if 1 card is chosen and not replaced? In this case, the probability of drawing a seven on the second draw is dependent on drawing a seven on the first draw. Now let’s calculate the probability of the 2 cards being drawn *without replacement*. This can be done with the Multiplication Rule.

Let \begin{align*}A = 1^{\text{st}}\end{align*}

Let \begin{align*}B = 2^{\text{nd}}\end{align*}

\begin{align*}& 4 \ \text{suits} \qquad 1 \ \text{seven} \ \text{per suit}\\
& \ \searrow \qquad \swarrow\\
\text{The total number of sevens in the deck} &= 4 \times 1=4.\end{align*}

\begin{align*}P(A) &= \frac{4}{52}\\
\\
& \qquad \qquad \ \ \text{Note: The total number of cards is}\\
P(B) &= \frac{3}{51} \ \swarrow \text{51 after choosing the first card if}\\
& \qquad \qquad \ \ \text{it is not replaced.}\\
\\
P(A \ \text{and} \ B) &= \frac{4}{52} \times \frac{3}{51} \ \text{or} \ P(A \cap B)=\frac{4}{52} \times \frac{3}{51}\\
\\
P(A \cap B) &= \frac{12}{2652}\\
\\
P(A \cap B) &= \frac{1}{221}\end{align*}

Notice in this example that the numerator and denominator decreased from \begin{align*}P(A)\end{align*}

#### Example B

A box contains 5 red marbles and 5 purple marbles. What is the probability of drawing 2 purple marbles and 1 red marble in succession *without replacement*?

On the first draw, the probability of drawing a purple marble is:

On the third draw, the probability of drawing a red marble is:

\begin{align*}P(\text{red}) =\frac{5}{8}\end{align*}

Therefore, the probability of drawing 2 purple marbles and 1 red marble is:

\begin{align*}P(1 \ \text{purple and 1 purple and 1 red}) &= P(1P \cap 1P \cap 1R)\\
&= P_1(\text{purple}) \times P_2(\text{purple}) \times P(\text{red})\\
&= \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8}\\
&= \frac{100}{720}\\
&= \frac{5}{36}\end{align*}

#### Example C

In Example B, what is the probability of first drawing all 5 red marbles in succession and then drawing all 5 purple marbles in succession *without replacement*?

The probability of first drawing all 5 red marbles in succession can be calculated as follows:

\begin{align*}P(1 \ \text{red and 1 red and 1 red and 1 red and 1 red}) &= P(1R \cap 1R \cap 1R \cap 1R \cap 1R)\\
&= P_1(\text{red}) \times P_2(\text{red}) \times P_3(\text{red}) \times P_4(\text{red}) \times P_5(\text{red})\\
&= \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}\\
&= \frac{120}{30240}\\
&= \frac{1}{252}\end{align*}

At this point, there are 5 marbles left, and all of them are purple. This means that with each remaining draw, the probability of drawing a purple marble is 1. Therefore, the probability of first drawing all 5 red marbles in succession and then drawing all 5 purple marbles in succession is \begin{align*}\frac{1}{252} \times 1 \times 1 \times 1 \times 1 \times 1 = \frac{1}{252}\end{align*}.

### Vocabulary

2 or more events whose outcomes influence each other are called ** dependent events**. With dependent events, the probability of occurrence of one event depends on the occurrence of the other.

### Guided Practice

Meg bought a box of chocolates that contained 18 caramels, 8 chocolate-covered cherries, and 14 truffles. What is the probability of Meg reaching into the box and pulling out a chocolate-covered cherry and then reaching in again and pulling out a caramel or a truffle? Assume that the chocolate-covered cherry is not replaced. What if the order were switched? In other words, what is the probability of Meg reaching into the box and pulling out a caramel or a truffle and then reaching in again and pulling out a chocolate-covered cherry *without replacement*?

**Answer:**

Since there is a total of \begin{align*}18+8+14=40\end{align*} chocolates in the box, the probability that Meg first pulls out a chocolate-covered cherry is \begin{align*}\frac{8}{40} = \frac{1}{5}\end{align*}. At this point, since the chocolate-covered cherry is not replaced, there are \begin{align*}40-1=39\end{align*} chocolates remaining in the box. Of those 39 chocolates, \begin{align*}18+14=32\end{align*} are either a caramel or a truffle. Therefore, the probability that Meg pulls out a caramel or a truffle after first pulling out a chocolate-covered cherry is \begin{align*}\frac{32}{39}\end{align*}.

Now that we know the probability of Meg first pulling out a chocolate-covered cherry and the probability of Meg then pulling out a caramel or a truffle, we can calculate the probability of both of these events happening. To do this, we can use the Multiplication Rule as follows:

\begin{align*}P(\text{chocolate-covered cherry}) \times P(\text{caramel or truffle}) = \frac{1}{5} \times \frac{32}{39} = \frac{32}{195}\end{align*}

If the order were switched, the probability that Meg first pulls out a caramel or a truffle is \begin{align*}\frac{32}{40} = \frac{4}{5}\end{align*}, and the probability that Meg then pulls out a chocolate-covered cherry is \begin{align*}\frac{8}{39}\end{align*}. Therefore, the probability of both events occurring can be calculated with the Multiplication Rule as follows:

\begin{align*}P(\text{caramel or truffle}) \times P(\text{chocolate-covered cherry}) = \frac{4}{5} \times \frac{8}{39} = \frac{32}{195}\end{align*}

Thus, the probability is the same, regardless of the order.

### Interactive Practice

### Practice

- Determine which of the following are examples of dependent events.
- Selecting a marble from a container and selecting a jack from a deck of cards.
- Rolling a number less than 4 on a die and rolling a number that is even on the same roll.
- Choosing a jack from a deck of cards and choosing another jack, without replacement.

- Determine which of the following are examples of dependent events.
- Selecting a book from the library and selecting a book that is a mystery novel.
- Rolling a 2 on a die and flipping a coin to get tails.
- Being lunchtime and eating a sandwich.

- Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? Assume that the first jelly bean is not replaced.
- For question 3, what if the order were reversed? In other words, what is the probability of Thomas reaching into the bag and pulling out a red jelly bean and then reaching in again and pulling out a blue or green jelly bean
*without replacement*? - What is the probability of drawing 2 face cards one after the other from a standard deck of cards
*without replacement*? - There are 3 quarters, 7 dimes, 13 nickels, and 27 pennies in Jonah's piggy bank. If Jonah chooses 2 of the coins at random one after the other, what is the probability that the first coin chosen is a nickel and the second coin chosen is a quarter? Assume that the first coin is not replaced.
- For question 6, what is the probability that neither of the 2 coins that Jonah chooses are dimes? Assume that the first coin is not replaced.
- Jenny bought a half-dozen doughnuts, and she plans to randomly select 1 doughnut each morning and eat it for breakfast until all the doughnuts are gone. If there are 3 glazed, 1 jelly, and 2 plain doughnuts, what is the probability that the last doughnut Jenny eats is a jelly doughnut?
- If the numbers 1 through 20 are each written on a slip of paper, and the slips of paper are placed in a hat, what is the probability that 2 slips of paper randomly chosen one after the other both have a prime number written on them? Assume that the first slip of paper is not replaced.
- Steve will draw 2 cards one after the other from a standard deck of cards
*without replacement*. What is the probability that his 2 cards will consist of a heart and a diamond?