 # 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),
the product of the outcome (xi) and the probability (pi)
for each case.

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

Sigma notation

## Example 1 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 For each toss, there are two outcomes:

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].

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

So x1 = 0.

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

(total) = 4 ways

So p1 = 1/4.

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.

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 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.