# Expected Value

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

## Formula

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 (*x*_{i}) and the probability (*p*_{i})

for each case.

So E(*X*) = *x*_{1}*p*_{1} + *x*_{2}*p*_{2} + *x*_{3}*p*_{3} + ... + *x*_{n}*p*_{n}.

Sigma notation

## Example 1

Case 1: Head

You get 3 points.

So *x*_{1} = +3.

The probability of getting a head is 1/2.

So *p*_{1} = 1/2.

Probability

Case 2: Tail

You lose 2 points.

So *x*_{2} = -2.

The probability of getting a tail is 1/2.

So *p*_{2} = 1/2.

*x*_{1} = +3, *p*_{1} = 1/2*x*_{2} = -2, *p*_{2} = 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

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 *x*_{i} and the probability *p*_{i}.

Case 1: 0 heads

So *x*_{1} = 0.

This is when [T, T] happens.

There's 1 way to happen this.

(total) = 4 ways

So *p*_{1} = 1/4.

Case 2: 1 head

So *x*_{2} = 1.

This is when [H, T] or [T, H] happens.

There are 2 ways to happen this.

(total) = 4 ways

So *p*_{2} = 2/4.

Case 3: 2 heads

So *x*_{3} = 2.

This is when [H, H] happens.

There's 1 way to happen this.

(total) = 4 ways

So *p*_{3} = 1/4.

*x*_{1} = 0, *p*_{1} = 1/4*x*_{2} = 1, *p*_{2} = 2/4*x*_{3} = 2, *p*_{3} = 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

The total area of the dart, *A*_{tot}, is*A*_{tot} = *π*⋅3^{2}.

So *A*_{tot} = 9*π*.

Area of a circle

Case 1: 30 points

So *x*_{1} = 30.

The area of the blue part, *A*_{1}, is*A*_{1} = *π*⋅1^{2}.

So *A*_{1} = *π*.

*A*_{1} = *π**A*_{tot} = 9*π*

So *p*_{1} = *π*/9*π*.

So *p*_{1} = 1/9.

Case 2: 20 points

So *x*_{2} = 20.

The area of the green part, *A*_{2}, is,

the circle with *r* = 2, *π*⋅2^{2}

minus,

the inner circle with *r* = 1, *π*⋅1^{2}.

So *A*_{2} = 3*π*.

*A*_{2} = 3*π**A*_{tot} = 9*π*

So *p*_{2} = 3*π*/9*π*.

So *p*_{2} = 3/9.

Case 3: 10 points

So *x*_{3} = 10.

The area of the brown part, *A*_{3}, is,

the circle with *r* = 3, *π*⋅3^{2}

minus,

the inner circle with *r* = 2, *π*⋅2^{2}.

So *A*_{3} = 5*π*.

*A*_{3} = 5*π**A*_{tot} = 9*π*

So *p*_{3} = 5*π*/9*π*.

So *p*_{3} = 5/9.

*x*_{1} = 30, *p*_{1} = 1/9*x*_{2} = 20, *p*_{2} = 3/9*x*_{3} = 10, *p*_{3} = 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.