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

karush

Well-known member
Jan 31, 2012
2,657
2020_05_01_11.11.18~2.jpg
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

Administrator
Staff member
Feb 24, 2012
13,736
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:

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

karush

Well-known member
Jan 31, 2012
2,657
oh I see
it's basically a series...

ummm where does 8 come from??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
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

Well-known member
Jan 31, 2012
2,657
oh..
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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

Well-known member
Jan 31, 2012
2,657
I'm very weak on this probability stuff
So the help was appreciated much