math expectation

ssmmss

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At the town fair, you can pay $5 to toss a ring at a set of bottles. If you get a “ringer” on the small mouth bottle, you win $35. If you get a “ringer” on the medium bottle, you win $10. If you get a “ringer” on the large bottle, you get your $5 fee back (that is, you break even). If you miss, you are out the $5 you paid to play. Joe is a good shot and his probability of getting a ringer on the small, medium, and large bottles is 10%, 10%, and 5%, respectively. The probability distribution of Joe’s winnings (accounting for the $5 that he paid to play) in a single game is given below.




X
-$5
$0
$10
$35
P
0.75
0.10
0.10
0.05


a. What is the math expectation of Joe’s winnings for a single game.

b. What is the math expectation of Joe’s winnings after 5 games.
 
Hello, ssmmss!

Some table entries are wrong and/or misplaced.

At the town fair, you can pay $5 to toss a ring at a set of bottles.
If you get a “ringer” on the small mouth bottle, you win $35.
If you get a “ringer” on the medium bottle, you win $10.
If you get a “ringer” on the large bottle, you win $5.
If you miss, you are out the $5 you paid to play.
Joe is a good shot and his probability of getting a ringer on the small, medium,
and large bottles is 10%, 10%, and 5%, respectively.

The probability distribution of Joe’s winnings (accounting for the $5 that he paid)
in a single game is given below.

. . \(\displaystyle \begin{array}{c|c|c|c|c|} & \text{Miss} & \text{L} & \text{M} & \text{S} \\ \hline
X & -\$5 & $0 & $5 & $30 \\ \hline
P & 0.75 & 0.05 & 0.10 & 0.10 \\ \hline \end{array}\)

a. What is Joe's expectation for a single game?
\(\displaystyle E \;=\;(0.75)(-$5) + (0.05)($0) + (0.10)($5) + (0.10)($30) \;=\;-$0.25 \)

Joe can expect to lose an average of \(\displaystyle 25\) cents per game.

b. What is Joe's expectation for 5 games?
For 5 games: .\(\displaystyle E \;=\;5(-$0.25) \;=\;-$1.25\)
 
Last edited by a moderator:
Some table entries are wrong and/or misplaced.

\(\displaystyle E \;=\;(0.75)(-$5) + (0.05)($0) + (0.10)($5) + (0.10)($30) \;=\;-$0.25 \)

Joe can expect to lose an average of \(\displaystyle 25\) cents per game.

For 5 games: .\(\displaystyle E \;=\;5(-$0.25) \;=\;-$1.25\)

Thank you soroban. The table entries are:

X -5$ $0 $10 $35
P .75 .1 .1 .05

I also calculated the variance and standard deviation but when I checked my results using an online calculator, my results were not quite matching.

This is what I did for var (x):

E(x) = -5(.75) + 0(.1) + 10(.1) + 35(.05) = -1
var (x) = (((-5-1)^2)*.75) + (((0-1)^2)*.1) + (((10-1)^2)*.1) + (((35-1)^2)*.05)
var (x) = 93

Std Deviation: square root of 93

Please advise.
 
Last edited by a moderator:
Thank you in advance, I figured out what my issue was.

Previously: var (x) = (((-5-1)^2)*.75) + (((0-1)^2)*.1) + (((10-1)^2)*.1) + (((35-1)^2)*.05)

Correction: var (x) = (((-5+1)^2)*.75) + (((0+1)^2)*.1) + (((10+1)^2)*.1) + (((35+1)^2)*.05)

Thanks again.
 
At the town fair, you can pay $5 to toss a ring at a set of bottles. If you get a “ringer” on the small mouth bottle, you win $35. If you get a “ringer” on the medium bottle, you win $10. If you get a “ringer” on the large bottle, you get your $5 fee back (that is, you break even). If you miss, you are out the $5 you paid to play. Joe is a good shot and his probability of getting a ringer on the small, medium, and large bottles is 10%, 10%, and 5%, respectively. The probability distribution of Joe’s winnings (accounting for the $5 that he paid to play) in a single game is given below.





X

-$5

$0

$10

$35

P

0.75

0.10

0.10

0.05


a. What is the math expectation of Joe’s winnings for a single game.

b. What is the math expectation of Joe’s winnings after 5 games.

Hi Soroban,

How would I find the cumulative distribution function?
 
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