A pair of honest dice is rolled once. Find the expected value of the sum of the two numbers rolled.

A Pair Of Honest Dice Is Rolled Once Find The Expected Value Of The Sum Of The Two Numbers Rolled 1 This question aims to find the expected value of the sum of two numbers in rolling a pair of dice. A common example of a  random trial is when a die is rolled. It is an act in which we can itemize all the attainable results that can be listed but the exact result on any provided part of the trial cannot be predicted accurately. In this case, a number will be allocated to every outcome known as the probability of the outcome to specify the likelihood of the occurrence of an event. A random trial is a process that generates a specific result that cannot be predicted with surety. A random experiment’s sample space is the set with all the potential results. Also, an event is said to be a subset of sample space. The product of the likelihood of an event with the number of times of occurrence of an event is said to be the expected value. The formula somewhat varies depending on the nature of the occurrences.

Expert Answer

Let $S$ be the sample space that contains the possible sum of numbers when two dice are rolled, then: $S=\{2,3,4,5,6,7,8,9,10,11,12\}$ Since a pair of dice is rolled, therefore the total number of samples is $36$. Let $x$ denote the sums in the sample space and let $p$ be their probabilities then:
$x$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$
$p$ $\dfrac{1}{36}$ $\dfrac{2}{36}$ $\dfrac{3}{36}$ $\dfrac{4}{36}$ $\dfrac{5}{36}$ $\dfrac{6}{36}$ $\dfrac{5}{36}$ $\dfrac{4}{36}$ $\dfrac{3}{36}$ $\dfrac{2}{36}$ $\dfrac{1}{36}$
$xp$ $\dfrac{2}{36}$ $\dfrac{6}{36}$ $\dfrac{12}{36}$ $\dfrac{20}{36}$ $\dfrac{30}{36}$ $\dfrac{42}{36}$ $\dfrac{40}{36}$ $\dfrac{36}{36}$ $\dfrac{30}{36}$ $\dfrac{22}{36}$ $\dfrac{12}{36}$
Now the formula for the expected value is: $E=\sum\limits_{i=1}^{11}x_ip_i$ $E=\dfrac{2}{36}+\dfrac{6}{36}+\dfrac{12}{36}+\dfrac{20}{36}+\dfrac{30}{36}+\dfrac{42}{36}+\dfrac{40}{36}+\dfrac{36}{36}+\dfrac{30}{36}+\dfrac{22}{36}+\dfrac{12}{36}$ $=\dfrac{2+6+12+20+30+30+42+40+36+30+22+12}{36}$ $=\dfrac{252}{36}$ $E=7$

Example 1

Harry rolls a fair die. Let $X$ be the event that the multiple of two occurs. Find the probability of $X$.


Let $S$ be the sample space, then the possible outcomes are: $S=\{1,2,3,4,5,6\}$ Number of sample points in the sample space $n(S)=6$ The required outcomes are $2,4,6$. Now, $P(X)=\dfrac{\text{Number of favorable outcomes}}{\text{Total Outcomes}}$ $P(X)=\dfrac{3}{6}$ $P(X)=\dfrac{1}{2}$ Hence, the probability of Harry getting a multiple of $2$ is $\dfrac{1}{2}$.

Example 2

A fair die is rolled $300$ times and there are $20$ chances of getting a $4$. Find the probability of getting a $4$.


Let $X$ be the probability of getting a $4$ then: $P(X)=\dfrac{20}{300}$ $=\dfrac{2}{30}$ $P(X)=\dfrac{1}{15}$

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