Expected Value – Solved Examples

Problem 1: Find the expected value of the outcome when a die is rolled. 

Solution: 

Consider X is a random variable that represents the value that comes when a die is rolled. 

X = {1, 2, 3, 4, 5, 6} 

Now, since the die is a fair die, the probability of getting each outcome is equal. That is [Tex]\frac{1}{6} [/Tex]

E(X) = [Tex]\sum x_{i}P(X = x_i) [/Tex]

⇒ E(X) = P(X = 1)(1) + P(X = 2)(2) + P(X = 3)(3) + P(X = 4)(4) + P(X = 5)(5) + P(X = 6)(6)

⇒ E(X) = [Tex]\frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) [/Tex]

⇒ E(X) = 3

Problem 2: A company makes phones. Out of 100 phones, one is faulty. For each phone, the company makes a profit of Rs2,000 and a loss of Rs 10,000 for the faulty phone. Find the expected profit. 

Solution: 

Let X be expected profit

E() = [Tex]\sum x_{i}P(X = x_i) [/Tex]

⇒ E(X) = P(X = Phone is working)(2000) + P(X = faulty phone)(10,000) 

⇒ E(X) = [Tex]\frac{49}{50}.2000 + \frac{1}{50}(-10,000) \\ = 1960 -200 \\ = 1760 [/Tex]

⇒ E(X) = 1760

Problem 3: In a three-time coin toss experiment. Find the expected number of heads. 

Solution: 

Let X be the number of heads obtained 

Sample Space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

E() = [Tex]\sum x_{i}P(X = x_i) [/Tex]

⇒ E(X) =P(X = 0)(0) + P(X = 1)(1) + P(X =2)(2) + P(X = 3)(3)

Finding out the probabilities, 

P(X = 0) = [Tex]\frac{1}{8} [/Tex]

P(X = 1) = [Tex]\frac{3}{8} [/Tex]

P(X = 2) = [Tex]\frac{3}{8} [/Tex]

P(X = 3) = [Tex]\frac{1}{8} [/Tex]

⇒ E(X) = P(X = 0)(0) + P(X = 1)(1) + P(X =2)(2) + P(X = 3)(3)

⇒ E(X) = [Tex]\frac{1}{8}. 0 + \frac{3}{8}.1 + \frac{3}{8}.2 + \frac{1}{8}.3 [/Tex]

⇒ E(X) =[Tex]\frac{12}{8} [/Tex]

⇒ E(X) = 1.5

Problem 4: Two friends are fishing in a pond that contains 5 trout and 5 sunfish. Each time they catch a fish, they release it back immediately. They made a bet. If the next three fishes friend A catches are all sunfish, then friend B will pay him Rs 10, otherwise, friend A will have to pay Rs 2 to B. Find the expected profit from the bet. 

Solution: 

Let X be a random variable denoting the profit from the bet. 

E(X) = P(A catches all three sunfish)(10) + P(A cannot catch all three sunfish)(-2)

Computing the probabilities, 

P(A catches all three sunfish) = [Tex]\frac{5}{10}.\frac{5}{10}.\frac{5}{10} = \frac{1}{8} [/Tex]

P(A doesn’t catch all three sunfish) = 1 – P(A catches all three sunfish)

                                                                = 1 – [Tex]\frac{1}{8} [/Tex]

                                                                = [Tex]\frac{7}{8} [/Tex]

Substituting the values computed above into the expectation equation, 

E(X) = P(A catches all three sunfish)(10) + P(A cannot catch all three sunfish)(-2)

⇒ E(X) = P(A catches all three sunfish)(10) + P(A cannot catch all three sunfish)(-2)

⇒ E(X) = [Tex]\frac{1}{8}.10 + \frac{7}{8}.(-2)  [/Tex]

⇒ E(X) = [Tex]\frac{-4}{8} = -\frac{1}{2} [/Tex]

Problem 5: Find the expected value of the outcome when a die is rolled. Given that the die is not fair, the probability of getting a 6 is 0.4 and the rest of the numbers are equally likely. 

Solution: 

Consider X is a random variable that represents the value that comes when a die is rolled. 

X = {1, 2, 3, 4, 5, 6} 

Now, since the die is not fair die, 

P(6) = 0.4 

P(1) = P(2) = P(3) = P(4) = P(5) = 0.12

E(X) = [Tex]\sum x_{i}P(X = x_i) [/Tex]

⇒ E(X) = P(X = 1)(1) + P(X = 2)(2) + P(X = 3)(3) + P(X = 4)(4) + P(X = 5)(5) + P(X = 6)(6)

⇒ E(X) = (0.12)(1) + (0.12)(2) + (0.12)(3) + (0.12)(4) + (0.12)(5) + (0.4)(6)

⇒ E(X) = (0.12)(1 + 2 + 3+ 4+ 5) + 2.4

⇒ E(X) = 1.8 + 2.4 

⇒E(X) = 4.2

Expected Value: Random variables are the functions that assign a probability to some outcomes in the sample space. They are very useful in the analysis of real-life random experiments which become complex. These variables take some outcomes from a sample space as input and assign some real numbers to it. The expectation is an important part of random variable analysis. It gives the average output of the random variable.