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# Independence versus Dependence

## Applications involving conditional probability and determining whether events are independent.

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Independence versus Dependence

### Independence Versus Dependence

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.

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*}

#### Calculating Probability

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

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*}

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

A student cannot take both math and art during the same period, so the events are not overlapping. If event A is having art first period and event B is having math first period, \begin{align*}P(A \ and \ B)=0\end{align*}. We want to find \begin{align*}P(A \ or \ B)\end{align*}.

\begin{align*}P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\end{align*}

\begin{align*}P(A \ or \ B) = \frac{40}{480} + \frac{96}{480} - 0\end{align*}

\begin{align*}P(A \ or \ B) = \frac{96}{480}\end{align*}

\begin{align*}P(A \ or \ B)=\frac{1}{5}=20\%\end{align*}

#### Finding 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*}.

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

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.

### Example

#### Example 1

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.

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.

### Review

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

1. Driving at night and falling asleep at the wheel.
2. Visiting the zoo and seeing a giraffe.
3. The next 2 cars you see are both red.
4. A coin tossed twice comes up heads both times.
5. Being dealt 4 aces in a hand of poker.
6. 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:

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

To view the Review answers, open this PDF file and look for section 13.8.

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### Vocabulary Language: English

conditional probability

The probability of a particular dependent event  given the outcome of the event on which it occurs.

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.

two way tables

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.

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