Expected Value

Expected Value

How to find the expected value from the given probability condition: formula, examples, and their solutions.

Formula

To find the expected value E(X), add up the product of the outcome and the probability for each case.

The expected value literally means
the expected value after an event.

To find the expected value E(X),
add up
the product of the outcome (xi) and the probability (pi)
for each case.

So E(X) = x1p1 + x2p2 + x3p3 + ... + xnpn.

Sigma notation

Example 1

A coin is tossed 1 time. If the head is shown, you get 3 points. If the tail is shown, you lose 2 points. Find the expected value of the points.

Case 1: Head

You get 3 points.
So x1 = +3.

The probability of getting a head is 1/2.
So p1 = 1/2.

Probability

Case 2: Tail

You lose 2 points.
So x2 = -2.

The probability of getting a tail is 1/2.
So p2 = 1/2.

x1 = +3, p1 = 1/2
x2 = -2, p2 = 1/2

So E(X) = 3⋅(1/2) + (-2)⋅(1/2).

The numerators of both fractions are 2.

So combine the fractions.

3 - 2 = 1

So E(X) = 1/2.

E(X) = 1/2 means
if a coin is tossed once,
you expect to get 1/2 points.

Example 2

A coin is tossed 2 times. Find the expected value of the number of the heads.

For each toss, there are two outcomes:
head or tail.

The coin is tossed 2 times.
And each toss is an independent event.

So (total) = 2⋅2
= 4 ways.

These 4 ways are
[H, H], [H, T], [T, H], and [T, T].

H: Head, T: Tail

For each case,
find the outcome xi and the probability pi.

Case 1: 0 heads

So x1 = 0.

This is when [T, T] happens.
There's 1 way to happen this.

(total) = 4 ways

So p1 = 1/4.

Case 2: 1 head

So x2 = 1.

This is when [H, T] or [T, H] happens.
There are 2 ways to happen this.

(total) = 4 ways

So p2 = 2/4.

Case 3: 2 heads

So x3 = 2.

This is when [H, H] happens.
There's 1 way to happen this.

(total) = 4 ways

So p3 = 1/4.

x1 = 0, p1 = 1/4
x2 = 1, p2 = 2/4
x3 = 2, p3 = 1/4

So E(X) = 0⋅(1/4) + 1⋅(2/4) + 2⋅(1/4).

0⋅(1/4) = 0

+1⋅(2/4) = +2/4

+2⋅(1/4) = +2/4

2/4 + 2/4 = 4/4

4/4 = 1

So E(X) = 1.

E(X) = 1 means
if a coin is tossed 2 times,
you expect to see 1 head.

Example 3

For the given dart, find the expected value of the points.

The total area of the dart, Atot, is
Atot = π⋅32.

So Atot = 9π.

Area of a circle

Case 1: 30 points

So x1 = 30.

The area of the blue part, A1, is
A1 = π⋅12.

So A1 = π.

A1 = π
Atot = 9π

So p1 = π/9π.

So p1 = 1/9.

Case 2: 20 points

So x2 = 20.

The area of the green part, A2, is,
the circle with r = 2, π⋅22
minus,
the inner circle with r = 1, π⋅12.

So A2 = 3π.

A2 = 3π
Atot = 9π

So p2 = 3π/9π.

So p2 = 3/9.

Case 3: 10 points

So x3 = 10.

The area of the brown part, A3, is,
the circle with r = 3, π⋅32
minus,
the inner circle with r = 2, π⋅22.

So A3 = 5π.

A3 = 5π
Atot = 9π

So p3 = 5π/9π.

So p3 = 5/9.

x1 = 30, p1 = 1/9
x2 = 20, p2 = 3/9
x3 = 10, p3 = 5/9

So E(X) = 30⋅(1/9) + 20⋅(3/9) + 10⋅(5/9).

The numerators of the fractions are all 9.

So combine the fractions.

30 + 60 + 50 = 140

So E(X) = 140/9.

E(X) = 140/9 means
if you throw a dart once,
you expect to get 140/9 points.