#### Objective

In this lesson, you will learn about calculating the probability that *all* of a series of ** independent events** will occur in a single experiment.

#### Concept

There is a classic example of probability studies involving a coin flip. Everyone knows that the probability of getting heads on a single flip is 50%, which means that every time you flip it, there is also a 50% probability of getting tails. The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?

After this lesson on the intersection of compound events, we’ll return to this question and see how it does (and doesn’t!) fit with the concept.

#### Watch This

http://youtu.be/xSc4oLA9e8o Khan Academy – Compound Probability of Independent Events

#### Guidance

It should make sense intuitively that the more specific or restricted you make the details of an event, the less probable it becomes for that event to occur. The concept of calculating the total probability of multiple events strung together is the same idea.

If I flip a coin once there are only two possible outcomes:

\begin{align*}H \ (heads) \ \text{or} \ T \ (tails)\end{align*}

If I flip the coin twice, there are four possibilities:

\begin{align*}H+T \ \text{or} \ H+H \ \text{or} \ T+H \ \text{or} \ T+T\end{align*}

We know there are a total of four possible outcomes from two coin flips: \begin{align*}HT\end{align*}, \begin{align*}HH\end{align*}, \begin{align*}TH\end{align*}, and \begin{align*}TT\end{align*}, and only one of them: \begin{align*}HH\end{align*}, results in the outcome we want to calculate. Using the simple probability formula, we get:

\begin{align*}P(HH)=\frac{1 \ outcome}{4 \ possible \ outcomes}=\frac{1}{4} \ or \ 25 \%\end{align*}

**Example A**

What is the probability of flipping a coin four times and getting tails all four times?

**Solution:** Create a table listing all of the possible outcomes:

Now we can look at the bottom row and see that there are a total of 16 possibilities, only one of which is four tails in a row. The probability, therefore, is:

\begin{align*}P(4 \ tails)=\frac{1 \ outcome}{16 \ outcomes}=\frac{1}{16}=6.25\%\end{align*}

**Example B**

What is the probability of rolling two even numbers in a row on a standard six-sided die?

**Solution:** Create a table listing all possible outcomes:

\begin{align*}P(two \ evens)=\frac{9 \ favorable \ outcomes}{36 \ total \ outcomes}=\frac{9}{36}=\frac{1}{4}\end{align*}

Reducing to:

\begin{align*}P(two \ evens)=\frac{1}{4} \ or \ 25\%\end{align*}

**Example C**

What is the probability of spinning two 2’s in a row OR two 4’s in a row on a spinner with the numbers 1-4?

**Solution:** Create a table listing all possible outcomes, and highlight the favorable ones:

Out of a total of 16 possible outcomes, only 2 fit our description, which gives us:

\begin{align*}P(2's \ or \ 4's )=\frac{2 \ \text{favorable} \ \text{outcomes}}{16 \ \text{possible} \ \text{outcomes}}=\frac{2}{16}=\frac{1}{8} \ or \ 12.5\%\end{align*}

**Concept Problem Revisited**

*The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?*

This is a very common example of something called the gambler’s fallacy. It is *not* a good example of calculating the intersection of compound events *because of the way it is worded*. The question as written is essentially asking about a single flip of the coin, which is always \begin{align*}\frac{50}{50}\end{align*}, because a coin has no memory.

From the standpoint of an example of what we have been studying in this chapter, the more useful, and dramatically more difficult question would be:

What is the probability of flipping a coin 100 times and getting heads every time?

See the difference? The first question assumes that 99 flips have already occurred and asks about the last flip, the second question asks about all 100 flips.

If you want to know the probability of flipping 100 heads in a row, you could either draw a *really* long chart of all of the possibilities (like the one in Example A, but much longer), or you could use the ** multiplication rule** that we will be learning in the next lesson. Check it out!

#### Vocabulary

An ** independent event** is an event whose outcome is not directly affected by another event. (a coin flip, for example)

A ** favorable outcome** is an outcome of an event that meets a set of initial specifications.

Two ** mutually exclusive** events cannot both occur at the same time, (e.g.

*both*heads

*and*tails on the same coin flip).

#### Guided Practice

1. What is the probability of pulling 1 red marble, replacing it, then pulling another red marble out of a bag containing 4 red and 2 white marbles?

2. What is the probability of a spinner landing on “2” and then a “3”, or “6” if there are 6 equally spaced points on the spinner?

3. What is the probability of pulling a red and then a black card at random from a standard deck (replacing the first card after drawing)?

4. What probability of picking a red and then a green marble from a bag with 5 red and 1 green marbles in it (replacing the first marble after the draw)?

5. What is the probability of shaking the hand of a student wearing red and then a student wearing blue if you randomly shake the hands of two people in a row in a room containing 3 students in blue and 2 in red?

**Solutions:**

1. Make a chart: \begin{align*}& \underline{\text{first pull:} \qquad \quad r \qquad \qquad r \qquad \quad \ r \qquad \quad \ \ r \qquad \quad \ w \qquad \quad w \quad \ \ }\\ & \text{second pull:} \ {\color{red}r r r r}w w \ \ {\color{red}r r r r} w w \ \ {\color{red}r r r r}w w \ \ {\color{red}r r r r}w w \ \ r r r r w w \ \ r r r r w w\end{align*}

The four sets of four red “\begin{align*}r\end{align*}’s” represent the favorable outcomes out of the total of 36, therefore \begin{align*}P(2 \ red)=\frac{16}{36}=\frac{4}{9} \ or \ 44.4\overline 4 \%\end{align*}

2. Make a chart: \begin{align*}& \underline{\text{first spin:} \qquad \qquad 1 \qquad \qquad \quad 2 \qquad\qquad \ 3 \qquad \qquad \ 4 \qquad \qquad \quad 5 \qquad \qquad \quad 6 \qquad }\\ & \text{second spin:} \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ {\color{red}3} \ 4 \ 5 \ {\color{red}6} \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6\end{align*}

The red numbers 3 and 6 represent the two favorable outcomes out of 36 total, therefore \begin{align*}P(2 \ and \ (3 \ or \ 6))=\frac{2}{36} \ or \ \frac{1}{18} \ or \ 5.5\overline{5} \%\end{align*}

3. There are 26 black and 26 red cards in the deck, so the probability on the *first* pull is \begin{align*}P(red)=\frac{26 \ \text{red cards}}{52 \ \text{total cards}}=\frac{26}{52}=\frac{1}{2} \ or \ 50 \%\end{align*} On the *second* pull, we again have a 50% chance of favorable outcome, but that 50% only applies to the half of the first pulls that were favorable. Therefore: \begin{align*}P(\text{red} \ then \ \text{black})=50\% \ of \ 50\%=25\%\end{align*}

4. Make a chart: \begin{align*}& \underline{\text{first pull:} \qquad \quad r \quad \qquad \ r \qquad \quad r \qquad \quad r \qquad \quad \ r \qquad \quad g \quad \ }\\
& \text{second pull:} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ \ r r r r r {\color{red}g} \ \ r r r r r g\end{align*} Of the 36 possible outcomes, only 5 fit the description of red the first time, and green the second time (noted by the red “\begin{align*}g\end{align*}’s”. **Therefore** \begin{align*}P(\text{red} \ then \ \text{green})=\frac{5}{36} \end{align*}

5. Make a chart:

So, out of the 25 possible handshake possibilities, 6 of them fit the requirements of red first, then blue:

\begin{align*}P(\text{red} \ then \ \text{blue})=\frac{6}{25}\end{align*}

#### Practice

Questions 1-6: Suppose you have an opaque bag filled with 4 red and 3 green balls. Assume that each time a ball is pulled from the bag, it is random, and the ball is replaced before another pull.

1. Create a chart of all possible outcomes of an experiment consisting of pulling one ball from the bag at random, noting the color and replacing it, then pulling another.

2. How many possible outcomes are there?

3. What is the probability of randomly pulling a red ball from the bag, returning it, and pulling agreen ball on your second pull?

4. What is the probability of randomly pulling a red ball both times?

5. What is the probability of pulling a green ball both times?

6. Is the probability of pulling a red followed by a green different than pulling a green followed by a red?

Questions 7 – 12: Suppose you have two standard dice, one red and one blue.

7. Construct a probability distribution table or diagram for an experiment consisting of one roll of the red die followed by one roll of the blue one.

8. How many possible outcomes are there?

9. Is there an apparent mathematical relationship between the number of sides on the dice and the number of possible outcomes?

10. What is the probability of rolling a 2 on the red die and a 1, 3, or 5 on the blue one?

11. What is the probability of rolling an even number on the red die and an odd on the blue one?

12. Do the probabilities of a particular outcome change based on which die is rolled first? Why or why not?

Questions 13 – 16: Suppose you have a spinner with 5 equally-spaced color sections: red, blue, green, yellow, and orange.

13. Construct a probability distribution detailing the possible outcomes of three consecutive spins. You may wish to use only the first letter, or a single color-coded hash mark, to represent each possibility, as there will be many of them.

14. How many possible outcomes are there?

15. Is there an apparent mathematical relationship between the number of sections on the spinner, the number of spins, and the number of possible outcomes? If so, what is the relationship?

16. What is the probability of spinning red, then green, and then orange?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.3.