Question

A fair coin is flipped three times. You win $5 every time the outcome is heads. Let the random variable X represent the total number of dollars you win.
(a) List the sample space.
(b) Determine the probability function of X.
======================
Answer: a) 2^3= 8 possibilities.
b) (P(heads 1st time)+ P(heads 2nd time) +P( heads 3rd time)) / 15

since each probability has 1/2 chance of occuring, the answer is 1/10

Answer 1

a) is correct

b) the Probability function is:
1/8: zero heads, $0 won
3/8: one heads, $5 won
3/8: two heads, $10 won
1/8: three heads, $15 won

Average winnings: 30/8 = $7.50

Answer 2

thank you drwls

Answer 3

Space:
t t t
t t h
t h t
t h h
h t t
h t h
h h t
h h h
So there are 8 possible patterns, agree with 2^3
there is One way to get 0 heads so 1/8 at 0
there are three ways to get 1 heads so 3/8 at 5
also
3/8 at 10
and
1/8 at 15
I get average = (1/8)(1*0+3*5+3*10+1*15) = 60/8
By the way, this is a binomial distribution problem
The probability of k heads in n tosses =
C(n,k)p(heads in a toss or .5))(1-p) which is also .5 for our coins
here C(n,k) = n!/[k!(n-k)!]

for n = 3 tosses
3! for example = 3*2*1 = 6
0! = 1 by convention
C(3,0) = 3!/[0!*3!) = 1
.5^0*.5^3 = 1/8
so p(0 heads) = 1*1/8 = 1/8 as we knew
C(3,1) = 6/[1!(2!)] = 6/2 = 3
.5^1*.5^2 = .125 = 1/8
so p(1 heads) = 3/8 as we knew
etc
now many of us know those coefficients by heart for small n, but you can also get the from making "Pascal's triangle
1 row zero
1 1 row 1
1 2 1 row 2
1 3 3 1 row 3 the one we want !
the zero and right elements are 1
between, each element is the sum of the two above it right and left.