# Thread: 2 probability theory questions!

1. ## 2 probability theory questions!

1)A small manufacturer sells 1 car each month with a probability of 0.3; it sells 2 cars each month with a probability of 0.1 and never sells more than 2 cars per month. Let X be the number of cars sold each month

-Find the mean and the variance of X
-If the monthly profit is 2(X^2)+3(X)+1, what is the expected montly profit?

2) In a game, Eric gives Heather three well-balanced quarters to flip. Heather will keep all the quarters that land in heads, and will give back the ones that land in teails. Yet, if all of them land in tails, Heather will pay two bucks to Eric. WHat is the expected value and the variance of Heather's net gain?

2. Originally Posted by Andreamet
1)A small manufacturer sells 1 car each month with a probability of 0.3; it sells 2 cars each month with a probability of 0.1 and never sells more than 2 cars per month. Let X be the number of cars sold each month

-Find the mean and the variance of X
-If the monthly profit is 2(X^2)+3(X)+1, what is the expected montly profit?

[snip]
Note that Pr(X = 0) = 0.6.

E(X) = (0)(0.6) + (1)(0.3) + (2)(0.1) = 0.5

E(X^2) = (0^2)(0.6) + (1^2)(0.3) + (2^2)(0.1) = 0.7

Variance = E(X^2) - [E(X)]^2 = 0.45

Let Y = 2X^2 + 3X + 1.

Then E(Y) = 2 E(X^2) + 3 E(X) + 1 = ......

3. Originally Posted by Andreamet
[snip]
2) In a game, Eric gives Heather three well-balanced quarters to flip. Heather will keep all the quarters that land in heads, and will give back the ones that land in teails. Yet, if all of them land in tails, Heather will pay two bucks to Eric. WHat is the expected value and the variance of Heather's net gain?
Let X = number of heads tossed.

Pr(X = 0) = 1/8. Gets -$2.00, that is loses$2.
Pr(X = 1) = 3/8. Gets $0.25. Pr(X = 2) = 3/8. Gets$0.50.
Pr(X = 3) = 1/8. Gets \$0.75.

E(X) = (-2.00)(1/8) + (0.25)(3/8) + (0.50)(3/8) + (0.75)(1/8) = ....

E(X^2) = ([-2.00]^2)(1/8) + (0.25^2)(3/8) + (0.50^2)(3/8) + (0.75^2)(1/8) = ....

Variance = E(X^2) - [E(X)]^2 = ......

4. thanks!l

5. Originally Posted by Andreamet
wouldn't the variance be 0.7 instead of 0.45?

woul
No. The variance is NOT the value of E(X^2).

Variance = E(X^2) - [E(X)]^2 = 0.7 - (0.5)^2 = 0.7 - 0.25 = 0.45.