Intro to Probability, please check my work.

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

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  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

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  2. thank you drwls

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  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
    3/8 at 10
    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
    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.

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