# Expected Value

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

## Formula

### Formula

E(X) = x_{1}p_{1} + x_{2}p_{2} + x_{3}p_{3} + ...

E(X): Expected value

x_{1}, x_{2}, x_{3}: Value for each case

p_{1}, p_{2}, p_{3}: Probability for each case

To find the expected value,

multiply the value (x_{i}) and the probability (p_{i})

for each case,

and add these products.

The expected value E(X) is also called

the mean.

## Example 1

### Example

### Solution

Case 1:

If you get a head,

you get 2 points.

x_{1} = 2

The probability of getting a head of a coin is

1/2.

So p_{1} = 1/2.

Then the expected value E(X) is,

x_{1}p_{1},

2⋅[1/2], ...

Case 2:

If you get a tail,

you lose 1 point.

x_{2} = -1

The probability of getting a tail of a coin is

1/2.

So p_{2} = 1/2.

So write, +x_{2}p_{2},

+(-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

### Example

### Solution

Case 1: Getting 1 point

x_{1} = 1

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

So p_{1} = 3/8.

So E(X) is,

x_{1}p_{1},

1⋅[3/8], ...

Case 2: Getting 2 points

x_{2} = 2

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

So p_{2} = 3/8.

So write, +x_{2}p_{2},

+2⋅[3/8].

Case 3: Getting 3 points

x_{3} = 3

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

So p_{3} = 2/8.

So write, +x_{3}p_{3},

+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.