Question

A fair die is rolled $10$times. Calculate the expected sum of the$10$ rolls.

Short answer

The expected sum of the$10$ rolls value are$35$.

Step-by-step solution

Given Information

Calculate the expected sum of the$10$ rolls.

Let random variable $X_i$represent the number on the face of the dice, after the roll.

Explanation

Since there are $6$numbers on a dice, the probability of getting any of the numbers is

$P\left(X_i=1\right)=\frac{1}{6},$

$P\left(X_i=2\right)=\frac{1}{6}$,

$P\left(X_i=3\right)=\frac{1}{6}$,

$P\left(X_i=4\right)=\frac{1}{6}$,

$P\left(X_i=5\right)=\frac{1}{6}$.

$P\left(X_i=6\right)=\frac{1}{6}$.

Expected sum of the 10 rolls

Let us find the expected sum of the $10$rolls.

$E\left(X_i\right)=1\times P\left(X_i=1\right)+2\times P\left(X_i=2\right)$

role="math" localid="1647251498640" $+3\times P\left(X_i=3\right)+4\times P\left(X_i=4\right)$

role="math" localid="1647251538454" $+5\times P\left(X_i=5\right)+6\times P\left(X_i=6\right)$

Substitute the value,

$=1\times \frac{1}{6}+2\times \frac{1}{6}$

$+3\times \frac{1}{6}+4\times \frac{1}{6}$

$+5\times \frac{1}{6}+6\times \frac{1}{6}$.

Multiply the value

Multiply the value,

$=\frac{1}{6}\times (1+2+3+4+5+6)$

$=\frac{21}{6}$

Divide

$=\frac{7}{2}$.

Substitute the value

Therefore, the expected sum of the $10$rolls is,

$E\left(X\right)=10\times E\left(X_i\right)$

Substitute,

$=10\times \frac{7}{2}$

$=35$.

Final answer

The expected sum of the$10$ rolls value are $35$.