MHB C Expected value for the total number of points awarded on any toss of coins

[karush]

https://www.physicsforums.com/attachments/9838
well not sure why we need 3 different coins other than confusion
also each toss at least 2 coins have to have the same face
frankly not sure how any of these choices work
didn't want to surf online better to stumble thru it here and learn it better

oh.. one nice thing about this new format... don't have to continuely log in

 

[MarkFL]

I would begin by listing the outcomes and their probabilities:

0 points: 1/8

3 points: 3/8

6 points: 3/8

9 points: 1/8

Then the expected value is the sum of the products of the points and probabilities associated with each outcome:

$$E[X]=0\cdot\frac{1}{8}+3\cdot\frac{3}{8}+6\cdot\frac{3}{8}+9\cdot\frac{1}{8}=\frac{9+18+9}{8}=\frac{9}{2}$$

 

[karush]

oh I see
it's basically a series...

ummm where does 8 come from??

 

[MarkFL]

Each coin has two possibilities, either heads or tails, and since there are 3 coins, the total number of outcomes is \(2^3=8\).

 

[karush]

oh..

 

[HallsofIvy]

Those 2^3= 8 outcomes are
HHH (worth 3(3)= 9 points)
HHT (worth 2(3)= 6 points)
HTH (woth 2(3)= 6 points)
HTT (worth 1(3)= 3 points)
THH (worth 2(3)= 6 points)
THT (worth 1(3)= 3 points)
TTH (worth 1(3)= 3 points)
TTT (worth 0(3)= 0 points)

 

[karush]

I'm very weak on this probability stuff
So the help was appreciated much