# 13.8: Independence versus Dependence

**At Grade**Created by: CK-12

**Practice**Independence versus Dependence

What if you were playing a game in which you picked a card from a standard deck and then rolled a die? How could you find the probability that you would pick an ace and roll a six? After completing this Concept, you'll be able to find the probabilities of independent events like these.

### Watch This

CK-12 Foundation: Independent Events

### Guidance

If the result of one event has no bearing on the probability of the second event, we call them **independent events**. For example, if you flip a coin 3 times and get heads 3 times, what is the probability that the next flip will result in tails? Many people think that the previous run of heads somehow influences the flip to make tails more likely, but in reality the previous flips have no bearing on the outcome of the new flip – how could they? The coin doesn’t have a brain or a memory.

(The idea that tails is more likely after a run of heads is called the Gambler’s Fallacy, and it probably arises from people getting confused about something called *prior probability*. Now that you’ve learned a little about probability, you know that getting three heads and one tail is a little more likely than getting four heads on four coin flips, so *before* you flip the coin, you’d expect that it’s more likely you’ll get three heads than four heads. But *after* you’ve already flipped the coin three times, the chances of getting heads on the first three flips don’t matter any more, because you *already got* those three heads; the probability of getting those first three heads has gone from 12.5% to 100%! So the only probability that still matters is the probability of getting heads on the one flip remaining, which is just the same as it always is on a single coin flip: 50%.)

Because one flip of the coin has no effect on the outcome of any other flips, each flip of the coin counts as an **independent event**.

To find the probability of multiple independent events happening together, we multiply the individual probabilities:

\begin{align*}\text{For independent events}: P(A \ and \ B) = P(A) \cdot P(B)\end{align*}

#### Example A

*Find the probability of rolling a 5 on a 6-sided die and getting heads if you flip a coin at the same time.*

**Solution**

Clearly the outcome of rolling a die has no effect on flipping a coin, so the two events are **independent**. So \begin{align*}P(5 \ \text{and heads}) = P(5) \cdot P(\text{heads}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.\end{align*}

#### Example B

*Out of the 480 students in a school, 40 have a birthday in February; also, 96 students take math in first period. Find the probability that a student picked at random will either have a birthday in February or take math in first period.*

**Solution**

A student’s birth month should have no effect on whether they take math in first period, so the two events are **independent**. However, there are also some students who will have both, so the events are also **overlapping**.

We’ll need to use the formula for overlapping events: \begin{align*}P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\end{align*}

But we’ll also need the one for independent events: \begin{align*}P(A \ and \ B) = P(A) \cdot P(B)\end{align*}

The probability of a February birthday is \begin{align*}P(\text{February}) = \frac{40}{480} = \frac{1}{12}\end{align*}.

The probability of a student taking math in \begin{align*}1^{st}\end{align*} period is \begin{align*}P(\text{math}) = \frac{96}{480} =\frac{1}{5}\end{align*}.

\begin{align*}P(\text{February and math}) &= P(\text{February}) \cdot P(\text{math})\\ &= \frac{1}{12} \times \frac{1}{5}\\ &= \frac{1}{60}.\end{align*}

\begin{align*}P(\text{February or math}) &= P(\text{February}) + P(\text{math}) - P(\text{February and math}) \frac{1}{12} + \frac{1}{5} - \frac{1}{60}\\ &= \frac{5+12-1}{60}.\end{align*}

The probability that a student picked at random will either have a birthday in February or take math in first period is \begin{align*}\frac{16}{60} = \frac{4}{15}\end{align*}, or **just under 27%**.

**Find the Probability of Dependent Events**

If the result of one event influences the probability of the second, we call them **dependent events**. For example, if you pick two cards from a deck, the chances of getting an ace on the first pick is \begin{align*}\frac{4}{52} = \frac{1}{13}\end{align*}. If you keep that ace and draw again, the chance of getting another ace on your second pick is less: there are now only 3 aces left in the deck (of 51 cards), so the chance of getting an ace is \begin{align*}\frac{3}{51} = \frac{1}{17}\end{align*}. To find the probability of getting two aces we multiply the two individual probabilities: \begin{align*}\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}\end{align*}.

#### Example C

*Three cards are picked from a standard 52 card deck. The cards are not replaced. Find the probability of picking 3 queens.*

**Solution**

There are 52 cards and 4 of them are queens, so the chance of getting a queen on the first pick is \begin{align*}\frac{4}{52} = \frac{1}{13}\end{align*}.

Assuming you get a queen on the first pick, there are 51 cards remaining of which 3 are queens, so the chance of getting a queen on the second pick is \begin{align*}\frac{3}{51} = \frac{1}{17}\end{align*}.

If you were successful on the second pick, there will be 50 cards remaining of which 2 are queens, so the chance of getting a queen on the third pick is \begin{align*}\frac{2}{50} = \frac{1}{25}\end{align*}.

The probability of picking 3 queens in a row is \begin{align*}\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} = \frac{1}{5525}\end{align*} or **1 in 5,525**.

Watch this video for help with the Examples above.

CK-12 Foundation: Independent Events

### Vocabulary

- If the result of one event has no bearing on the probability of the second event, we call them
**independent events**.

- If the result of one event influences the probability of the second, we call them
**dependent events**.

### Guided Practice

*100 raffle tickets were sold and Peter bought 4 of them. There are 3 prizes, and winners are selected randomly from a hat containing all the numbers. Find the probability that Peter wins all 3 prizes.*

**Solution**

For the first draw, Peter’s numbers account for 4 tickets out of 100: \begin{align*}\frac{4}{100} = \frac{1}{25}\end{align*}

For the second draw, Peter’s remaining numbers (assuming he won the first draw) account for 3 tickets out of 99: \begin{align*}\frac{3}{99} = \frac{1}{33}\end{align*}

For the first draw, Peter’s remaining numbers (assuming he won the first two draws) account for 2 tickets out of 98: \begin{align*}\frac{2}{98} = \frac{1}{49}\end{align*}

The probability that Peter wins all 3 prizes is \begin{align*}\frac{1}{25} \times \frac{1}{33} \times \frac{1}{49} = \frac{1}{40425}\end{align*} or **1 in 40, 425**.

### Practice

For 1-6, determine whether the events are dependent or independent.

- Driving at night and falling asleep at the wheel.
- Visiting the zoo and seeing a giraffe.
- The next 2 cars you see are both red.
- A coin tossed twice comes up heads both times.
- Being dealt 4 aces in a hand of poker.
- It is your birthday and it is a windy day.

For 7-10, a bag contains 10 colored marbles – 4 red, 4 blue and 2 green. Calculate the probability of:

- Removing 2 green marbles in a row if you replace the marble each time.
- Removing 2 green marbles in a row if you do not replace the marble each time.
- Removing 3 marbles without replacing and getting all blue.
- Removing 4 marbles without replacing and getting
**exactly**3 blue.

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

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Contingency tables are sometimes called two-way tables because they are organized with the outputs of one variable across the top, and another down the side.### Image Attributions

Here you'll learn how to find the probability of two events (called independent events) in which one event has no bearing on the probability of the second event. You'll also learn how to find the probability of two events (called dependent events) in which one event does have a bearing on the probability of the second event.