9-8 Mathematical Expectation

Lesson 9-8: Mathematical Expectations

In this lesson, we will learn about how you can calculate an expected value based on outcomes for each event in a random experiment. Basically, we will put our knowledge of probability to good use.

The example the book gives is of a carnival game of rolling a die.
If you...
roll a 6-- you win 100 points
roll a 2 or 4–- you win 10 points
roll an odd number–- lose 50 points

So in chart form,

Number on die ------Points you win
1 --------------------(-50)
2--------------------(10)
3 --------------------(-50)
4 --------------------(10)
5 --------------------(-50)
6 --------------------(100)

Strategy No. 1: To find the average point winning from the rolling-a-die-game...
You add up all the possible point outcomes and divide by the number of rolls!

(-50+10-50+10-50+100)/(6rolls) It simplifies too..
-30points/6 rollsor -5points/roll

What does it mean???---This answer means that you will on average lose 5 points per roll of the die.

Strategy No. 2

(-50 + 10 + -50 + 10 + -50 + 100) /divided by 6 rolls.
--- (The different outcomes of pts are in numerator, and the number of rolls/trials is in denominator)

(3*(-50) + 2*(10) + 1*(100)) /divided by 6 rolls --- you can "combine like terms"

3/6 * (-50) + 2/6 *(10) + 1/6 * (100) ---combine into individual fractions so that you can see what the probabilities for each outcome is!!!

probability of losing 50 pts is 3/6
probability of winning 10 pts is 2/6
probability of winning 100 pts is 1/6

Strategy No. 3 is the Algebraic Formula:

Mathematical Expectation = (Probability of Event A) * (Outcome of Event A) + (probability of event B) * (outcome of event B) +...

In the die-rolling example:avg winning = [(probability of rolling a 6) *(100 points)] + [(probability of rolling an odd number) * (-50 points) ] + [(probability of rolling a 2 or 4) * (10 points)]

PROBLEM #2:
The 2nd example that the book gives is more difficult because the probability involves "binomial probability distribution" from section 9-7...which means you have to use the equation from pg. 387 of the text book...

P(x) = nCx * a^n-x * b^x

b is the probability that the event occurs
a is the probability that the event does not occur
x is the number of times the event occurs in n repetitions

NOTE: use binomial probability distribution when there are only 2 possible outcomes...like getting a basketball shot---you either get it in or not. Whereas, you wouldn't use this with the die problem because there is an equal probability of rolling each of the 6 numbers on the die.

Basketball Toss Problem: You get 3 tries to shoot basketball hoops.

Number of Baskets------- Points (outcome)
0-------------------------(-50)
1 -------------------------(-45)
2 -------------------------(10)
3 -------------------------(200)

Probability of making the shot = 30% or .3
Probability of not making the shot = 70% or .7

To Solve!!!Now, find the probability of making x baskets...USE binomial probability distribution equation!!!



P(0) = 3C0 * .7^3 * .3^0 = .343 probability of getting 0 shots

P(1) = 3C1 * .7^2 * .3^1 =.441 prob. of getting 1 shot

P(2) = 3C2 * .7^1 * .3^2 = .189 prob. of getting 2 shots

P(3) = 3C3 * .7^0 * .3^3 = .027 prob. of getting 3 shots

(**note: the probabilities of all the outcomes add up to 1)


Now, just plug it into the Mathematical Expectation Equation!
mathematical expectation (E) = (prob. of 1st event * outcome of 1st event) + ...2nd...3rd....

E = .343 * (-50) + .441 * (-45) + .189 * (10) + .027 * (200)

E= about -30 points per game

ALL DONe!

Sites for more information on mathematical expectation:
http://www.wiu.edu/users/mfmk/Math101/Probability/PExpect.html
http://engrwww.usask.ca/classes/GE/210/CE%20225%20Webpage%20files/LECTURES/Lecture%208/Updated%20Lecture%208.pdf


Personalized Part: I love Miró. In my 3rd grade class with Mrs. Schroeder, we learned about Miró and all tried to copy his style...



























Reminder!! Nick, you are NEXT!