# Expected Value

How to find the expected value of an event: formula, 2 examples, and their solutions.

## Formula

### Formula

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

E(X): Expected value
x1, x2, x3: Value for each case
p1, p2, p3: Probability for each case

To find the expected value,
multiply the value (xi) and the probability (pi)
for each case,

The expected value E(X) is also called
the mean.

## Example 1

### Solution

Case 1:
you get 2 points.

x1 = 2

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

Then the expected value E(X) is,
x1p1,
2⋅[1/2], ...

Case 2:
If you get a tail,
you lose 1 point.

x2 = -1

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

So write, +x2p2,
+(-1)⋅[1/2].

So E(X) = 2⋅[1/2] + (-1)⋅[1/2].

2⋅1 = 2
+(-1)⋅1 = -1

2 - 1 = 1

So E(X) = 1/2.

This means
if a coin is tossed once,
you expect to get 1/2 points.

## Example 2

### Solution

Case 1: Getting 1 point

x1 = 1

The 1 point area is 3/8 of the whole spinner.
So p1 = 3/8.

So E(X) is,
x1p1,
1⋅[3/8], ...

Case 2: Getting 2 points

x2 = 2

The 2 points area is 3/8 of the whole spinner.
So p2 = 3/8.

So write, +x2p2,
+2⋅[3/8].

Case 3: Getting 3 points

x3 = 3

The 3 points area is 2/8 of the whole spinner.
So p3 = 2/8.

So write, +x3p3,
+3⋅[2/8].

So
E(X) = 1⋅[3/8] + 2⋅[3/8] + 3⋅[2/8].

1⋅3 = 3
+2⋅3 = +6
+3⋅2 = +6

3 + 6 + 6 = 15

So E(X) = 15/8.

This means
if you spin the arrow once,
you expect to get 15/8 points.