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1. ## Binomial Problem

Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

A carnival Sharpshooter game charges \$25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back \$10. For each additional bullseye over 5 he gets back an additional \$5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest \$1)?

Ok they set up the Expected prize as

E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

why is 5.p(4) subtracted???

2. I haven't looked at it too long but these problems/solutions tend to be given in "layers" so you subtract to account for the layer where you get 0. Also, did you mean to have an "=" somewhere in there?

3. Originally Posted by hershey79
Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

A carnival Sharpshooter game charges \$25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back \$10. For each additional bullseye over 5 he gets back an additional \$5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest \$1)?

Ok they set up the Expected prize as

E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

why is 5.p(4) subtracted???
I'm not sure how you got that. From what I recall,

Let x = number of bullseye

E(gains) = (-25)p(x=0,x=1,x=2,x=3,x=4) + (-15)p(x=5) + (-10)p(x=6) + ... + (80)p(x=25)

Solving for each p(x) and you get your expected gain

4. Originally Posted by hershey79
Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

A carnival Sharpshooter game charges \$25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back \$10. For each additional bullseye over 5 he gets back an additional \$5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest \$1)?

Ok they set up the Expected prize as

E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

why is 5.p(4) subtracted???

X - # of bullseye ~ Binomial ( n=25, p=.2)

X = 0 1 2 3 4 5 6 7 . . . 25
5X-15 = -15 -10 -5 0 5 10 15 20 . . . 110

The question asks for E[5X-15] but not including those X=0,1,2,3,4

Bottomline: Look at the signs of 5X-15 on those X's

You should be able to figure out. :smiloe:

5. Originally Posted by KHC831
I'm not sure how you got that. From what I recall,

Let x = number of bullseye

E(gains) = (-25)p(x=0,x=1,x=2,x=3,x=4) + (-15)p(x=5) + (-10)p(x=6) + ... + (80)p(x=25)

Solving for each p(x) and you get your expected gain
Possible but good luck on doing this

6. ''' thanks for the inputs!

I will take again a look at those signs.

Hershey79

7. Originally Posted by hershey79
''' thanks for the inputs!

I will take again a look at those signs.

Hershey79
Note: E[X-15] is the expected prize not yet the answer on the question which asks for the expected gain.