Expected Value
How to find the expected value of an event: formula, 2 examples, and their solutions.
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,
and add these products.
The expected value E(X) is also called
the mean.
Example
E(X) = 1/2
This means
if a coin is tossed once,
you expect to get 1/2 points.
Case 1:
If you get a head,
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
E(X) = 15/8
This means
if you spin the arrow once,
you expect to get 15/8 points.
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.