Question

Question:

(a) Define the expected value of a random variable \(X\).

(b) What is the expected value of the random variable \(X\) that assigns to a roll of two dice the larger number that appears on the two dice?

Short answer

Answer

(a) The resultant answer is\(E(X) = \sum\limits_{x \in S} x P(X = x)\).

(b) The resultant answer is \(\frac{{161}}{{36}} \approx 4.4722\).

Step-by-step solution

Given data

The given datais two dice.

Simplify by using the concept of random variable

(a)

The expected value is the sum of the product of each possibility in the sample space with its probability \(P(x)\) :\(E(X) = \sum\limits_{x \in S} x P(X = x)\).

Find the possible values of the dice 

(b)

Let \(X\) represent the outcome of the first die and \(Y\) the outcome of the second die.

In the table below, we note that there are 36 outcomes of the two dice and thus each outcome has 1 chance in 36 of occurring:

\(P(X = x,Y = y) = \frac{1}{{36}}\)

The expected value is the sum of the product of each possibility \(x\) in the sample space with its probability \(P(x)\) :

\(\begin{aligned}{}E(\max (X,Y)) &= \sum\limits_{x \in S} {\sum\limits_{y \in S} {\max } } (x,y)P(X = x,Y &= y)\\E(\max (X,Y)) &= \frac{1}{{36}}\sum\limits_{x \in S} {\sum\limits_{y \in S} {\max } } (x,y)\\E(\max (X,Y)) &= \frac{1}{{36}}(1 \cdot 1 + 3 \cdot 2 + 5 \cdot 3 + 7 \cdot 4 + 9 \cdot 5 + 11 \cdot 6)\\E(\max (X,Y)) &= \frac{{161}}{{36}}\end{aligned}\)

\(E(\max (X,Y)) \approx 4.4722\)