# Rolling Dice: You roll a die with 6 faces 10 times. What is the expected number of faces that...

## Question:

Rolling Dice: You roll a die with 6 faces 10 times.

What is the expected number of faces that appear twice?

## Using a Binomial Random Variable to determine Expected Value :

A simple way to determine whether a random variable follows BInomial distribution is to ask yourself the following questions:

- Is each trial independent?
- Is the outcome of each trial is either a success or a failure?
- Does the probability of success remain the same in every trial?
- Are there a fixed number of trials?

If you can answer yes to all these questions, then the variable follows Binomial distribution. With this, we can use the following formula to approximate the number of times a certain success may occur:

{eq}P(X=k)= \binom{n}{k} [P(success)]^k [P(failure)]^{(n-k)} {/eq}

Where k represents the number of times a success occurs!

## Answer and Explanation:

In this problem, we can recognize the answer to the 4 questions (from the context section) to determine if a random variable follows a binomial distribution. Every time the die is thrown, the chance of a certain face landing in one trial does not affect the chance of it landing in another trial. This means each trial is independent. In each trial, a certain number on the die lands or it does not with probabilities p and 1-p respectively. This means the outcome is either a success or a failure. The chance of a face landing stays the same each time the die is rolled. Lastly, the question clearly states that the die will be rolled for a set number of trials. This means the probability of a face appearing twice can be determined using the Binomial Distribution formula. Before we set it up, we must recognize the following:

{eq}n = 10 trials {/eq}

{eq}k = 2 success {/eq}

{eq}P (success) = P (face landing) = 1/6 {/eq}

and,

{eq}P (failure) = 1- P (success) = 1 - 1/6 = 5/6 {/eq}

Thus, the Binomial formula can be set up as follows:

{eq}P (X) = \binom{10}{2} (1/6)^2 (5/6)^8 = 0.291 {/eq}

After determining this probability, we can set up the following probability distribution table:

Number of sides (X) | P (X) |
---|---|

6 | 0.291 |

From this we can determine the expected number of faces that appear twice as follows:

{eq}E(X) = (X) P(X) = (6) (0.291) = 1.746 {/eq}

Thus, on average a face will appear twice about 1.746 times in 10 rolls.

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from High School Algebra I: Help and Review

Chapter 24 / Lesson 5