# Normal Average versus Weighted Average

#### B1966J

##### New Member
I'm taking a self-study course from "The Great Courses" called "Joy of Numbers". I'm on the lesson "Joy of Mathematical Games" and am confused by the professor's seemingly interchangeable use of the terms "weighted probabilities" and "weighted averages".

The example given is the Law of Total Probability where:

If event B has 3 possible outcomes B1, B2, B3, then

P(A) = P(A|B1)P(B1)+P(A|B2)P(B2)+P(A|B3)P(B3)

The professor calls the result the "Weighted Average" of the event's probabilities.

So, if this is an average, why isn't everything divided by 3?

e.g. P(A) = (P(A|B1)P(B1)+P(A|B2)P(B2)+P(A|B3)P(B3))/3

When I work through a real life scenario like -
A horse (A|B) wins a race 60% of the time when it rains, 50% of the time when it snows and 20% of the time when it's clear. And, the probability of rain (B) is 80%, snow is 19% and clear skies is 1%, then the percentage probability of the horse winning any given race is:

P(horse win) = (0.60)(0.80)+(0.50)(0.19)+(0.20)(0.01) = 0.577

Intuitively that makes sense because you would expect that with a high probability of rain the horse would come close to 60% probability of winning.

But I'm having trouble reconciling the term "average" with something that doesn't take the number of terms in the equation into consideration.

Can anyone help me with this?

#### rogojel

##### TS Contributor
hi,
a normal average with 3 terms can be thought of as x1*1/3+x2*1/3+x3*1/3. A general weighted average will be something like x1*a1+x2*a2+x3*a3 with the condition that a1+a2+a3=1.

In your case, because P(B1)+P(B2)+P(B3)=1 this is a weighted average .

I hope this helps

regards