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# Counting Events

## And' means multiply

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Multiplication Rule

Finding the probability of getting two or three heads in a row when flipping a fair coin is straightforward enough by building a frequency table. However, the process becomes somewhat unwieldy when the experiment is more complex, such as calculating the probability of pulling 3 queens in a row from a standard deck of cards. Building a frequency table for all 52 cards would be time consuming at best.

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There must be an easier way, right?

### Multiplication Rule

If I bet someone $1 that I can roll a standard die and get a 6, and then do roll the die and get a 6, I would probably get my$1, along with a clap on the back and a “congratulations!” from my friends. However, if I bet someone \$20 that I can roll 20 times and get 6 every time, and then I do just that, I would probably be dealing with a very angry group of people who want to know how I cheated! That’s because it would be, at best, very improbable that I could get a series of 20 6’s in a row under normal circumstances. Let’s see if we can find out just how improbable.

Let’s start with the probability of just one roll of a 6:

Since there are 6 sides, the probability is:

\begin{align*}P(6)=\frac{1 \ outcome}{6 \ possible \ outcomes}=\frac{1}{6}\ or \ 16.7\%\end{align*}

For two 6’s in a row:

If we create a table of the possible rolls where the only total outcome yielding two 6’s is highlighted in blue, we get:

KEY

With this already pretty unwieldy table, we can see that there are 36 possible outcomes of rolling a standard die twice. Therefore, the probability of rolling two 6’s in a row is:

\begin{align*}P(two \ 6's)=\frac{1 \ outcome}{36 \ possible \ outcomes}=\frac{1}{36}\ or \ 2.8\%\end{align*}

It should be apparent that things only get crazier from here, since calculating 3 6’s this way would require another row of 6 possibilities for each of the 36 outcomes of the 2nd roll! However, look at the difference between the two probabilities:

\begin{align*}P(one \ 6)=\frac{1}{6}\ and \ P(two \ 6's)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}\end{align*}

The fact that the probability of getting two 6’s is \begin{align*}\frac{1}{6^{th}}\end{align*} of \begin{align*}\frac{1}{6}\end{align*} is no coincidence, of course. In fact, to understand how to calculate more complex intersections of independent compound probabilities, it may help to remember something you likely learned when practicing word problems:

To translate English to math, the word of and the multiplication sign · or × mean the same thing.

Since the probability of getting two 6’s in a row is \begin{align*}\frac{1}{6^{th}} \ of \ \frac{1}{6^{th}}\end{align*} we can say:

\begin{align*}P(two \ 6's)=\frac{1}{6} \ of \ \frac{1}{6}=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}\end{align*}

This is an example of the multiplication rule of compound probability:

\begin{align*}P(total)=P(1st \ outcome)\times P(2nd \ outcome)\ldots \times \ P(last \ outcome)\end{align*}

Now we can actually calculate the probability of 20 6’s in a row:

\begin{align*}P(20 \ 6's)=\frac{1}{6}\times \frac{1}{6} \ldots \frac{1}{6}=\left(\frac{1}{6}\right)^{20}=\frac{1}{3,656,158,440,062,976}\end{align*}

or approximately one in three and one-half quadrillion, which I would consider not good odds!

Which also illustrates the practical impossibility of solving such a question with a frequency table, since it would take approximately 116,000,000 years just to write out the 6th row at one number per second! (not to mention the 1 trillion sheets of paper\begin{align*}\ldots\end{align*})

#### Finding Theoretical Probability

1. What would be the theoretical probability of randomly pulling a queen from a deck of 52 cards, putting it back, randomly pulling a queen again, and so on until you have pulled 5 queens in a row?

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The theoretical probability of pulling a single queen from a standard deck is:

\begin{align*}P(queen)=\frac{4 \ \text{queens}}{52 \ \text{cards}}=\frac{1}{13} \ or \ 7.7\%\end{align*}

If we use the multiplication rule for five pulls, we get:

\begin{align*}\frac{1}{13}\times \frac{1}{13}\times \frac{1}{13}\times \frac{1}{13}\times \frac{1}{13}= \frac{1}{13^5}=\frac{1}{371293}\end{align*}

2. What is the theoretical probability of rolling a 1, 2, 3, 4, 5, and then 6, in order, on six successive rolls of a standard die?

The probability of rolling any single number on a standard die is \begin{align*}\frac{1}{6}\end{align*}. Use the multiplication rule:

\begin{align*}P(1-6)=\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{6^6}=\frac{1}{46656}\end{align*}

3. What is the theoretical probability that you might deal the King of Hearts, Jack of Diamonds, and then any Ace, in order, from a standard deck of cards, assuming you replace each card after drawing?

Let’s start by evaluating the individual probabilities.

• \begin{align*}P(first \ card)=\frac{\text{1 King of Hearts}}{\text{52 total cards}}=\frac{1}{52}\end{align*}
• \begin{align*}P(second \ card)=\frac{\text{1 Jack of Diamonds}}{\text{52 total cards}}=\frac{1}{52}\end{align*}
• \begin{align*}P(third \ card)=\frac{\text{4 Aces}}{\text{52 total cards}}=\frac{4}{52}=\frac{1}{13}\end{align*}

Now we can use the multiplication rule:

\begin{align*}P(King \ of \ Hearts, \ Jack \ of \ Diamonds, \ Ace)=\frac{1}{52}\times \frac{1}{52} \times \frac{1}{13}=\frac{1}{35152}\end{align*}

#### Earlier Problem Revisited

Finding the probability of getting two or three heads in a row when flipping a fair coin is straightforward enough by building a frequency table. However, the process becomes somewhat unwieldy when the experiment is more complex, such as calculating the probability of pulling 3 queens in a row from a standard deck of cards. Building a frequency table for all  cards would be time consuming at best.

There must be an easier way, right?

Of course, now you know this isn’t even really a question anymore, the multiplication rule makes this question pretty easy:

\begin{align*}P(3 \ queens)=\frac{1}{52}\times \frac{1}{51} \times\frac{1}{50}=\frac{1}{132600}\end{align*}

### Examples

#### Example 1

What is the probability of rolling an odd number, followed by an even number, followed by a prime number, on three successive rolls of a 20-sided die?

Apply the multiplication rule: \begin{align*}P(total) =P(\text{case} \ 1)\times P(\text{case} \ 2)\ldots \times P(\text{case} \ n)\end{align*}

\begin{align*}P(odd, \ even, \ prime) &=P(odd) \times P(even) \times P(prime)\\ P(odd, \ even, \ prime) &=\frac{10}{20} \times \frac{10}{20} \times \frac{8}{20}=\frac{800}{8000}=\frac{1}{10} \ or \ 0.1 \ or \ 10\%\\ P(odd, \ even,\ prime)&=10\% \end{align*}

#### Example 2

What is the probability of pulling a heart, replacing it, pulling a club, replacing it, pulling a diamond, replacing it, then pulling a spade, all from a standard deck?.

We can apply the multiplication rule here also:

\begin{align*}P(heart,\ club,\ diamond,\ spade) &=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{256} \ or \ 0.00391 \ or \ .4\% \\ P(heart, \ club, \ diamond, \ spade) &=0.4\% \end{align*}

#### Example 3

What is the probability of flipping a coin ten times in a row, and getting heads every time?

Since we are looking for the same outcome from the same experiment repeated ten times, we can 'shortcut' the multiplication rule by an exponent:

\begin{align*}P(heads \ ten \ times) &=\left(\frac{1}{2}\right)^{10}=\frac{1^{10}}{2^{10}}=\frac{1}{1024} \ or \ 0.001 \ or \ 0.1\% \\ P(heads \ ten \ times) &=0.1\% \end{align*}

#### Example 4

What is the probability of spinning red 5 times in a row on a spinner with 6 equally spaced color segments, only one of which is red?

This one is similar to the last, in that we are looking for the same outcome of the same experiment, multiple times (5 times, in this case).

\begin{align*}P(red \ five \ times) &=\left(\frac{1}{6}\right)^5=\frac{1^5}{6^5}=\frac{1}{7776} \ or \ 0.0001 \ or \ 0.01 \% \\ P(red \ five \ times) &=0.01\% \end{align*}

### Review

1. What are independent events?

2. What is the multiplication rule?

Questions 3-7: Suppose you have an opaque bag filled with 6 red, 4 green, 7 blue and 5 purple balls.

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

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

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

6. What is the probability of randomly pulling a blue ball from the bag, returning it, and pulling ared ball on your second pull?

7. What is the probability of randomly pulling a purple ball from the bag, returning it, and pulling ablue ball on your second pull?

Questions 8 – 12: Suppose you have two standard dice, one red and one blue.

8. What is the probability of rolling a 3 on the red die and a 5 on the blue one?

9. What is the probability of rolling a 3 or 4 on the red die and a 5 on the blue one?

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

11. What is the probability of rolling a 6 on the red die and an odd number on the blue one?

12. What is the probability of rolling a 1 on the red die and prime number on the blue one?

Questions 13 – 16: Suppose you are dealing with a standard deck of cards, calculate the probability of each outcome as described, assuming you replace each card after drawing it.

13. Pulling a queen, then any club, then any red card.

14. Pulling the Ace of Spades, then a red 6, then any king.

15. Pulling any face card three times in a row.

16. Pulling a face card, then an ace, then the 5 of clubs.

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### Vocabulary Language: English

multiplication rule of probability

The multiplication rule of probability states that, for independent events: P(total) = P(Case 1) x P(Case 2) x ... P(Case n).