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.
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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
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
thank you drwls
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.
To determine the sample space of flipping a fair coin three times, we need to consider all possible outcomes. In this case, each coin flip can result in either heads (H) or tails (T), giving us two possibilities for each flip. Since we are flipping the coin three times, we need to consider all the combinations of heads and tails.
(a) To list the sample space, we can consider each flip as a distinct event and list all the possible outcomes. Let's denote heads as H and tails as T. The sample space consists of the following eight possibilities:
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
(b) To determine the probability function of the random variable X, which represents the total number of dollars you win, we need to calculate the probability of each possible outcome.
Since each flip of a fair coin has an equal chance of resulting in either heads or tails, the probability of getting heads on any individual flip is 1/2, and the probability of getting tails is also 1/2.
Since we win $5 for every heads outcome, the value of X will be the sum of the number of heads in the three flips.
To calculate the probability function of X, we need to determine the probabilities of obtaining different values for X. Since the total number of heads can be 0, 1, 2, or 3, we need to calculate the probability of each possibility.
For X = 0: The probability of getting no heads is only achieved when we get tails on all three flips. The probability is (1/2) * (1/2) * (1/2) = 1/8.
For X = 1: The probability of getting one heads can be achieved in three ways: HTH, THH, or HHT. The probability is (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) = 3/8.
For X = 2: The probability of getting two heads can also be achieved in three ways: HH T, T HH, or H TT. The probability is (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) = 3/8.
For X = 3: The probability of getting three heads is only achieved when we get heads on all three flips. The probability is (1/2) * (1/2) * (1/2) = 1/8.
To calculate the probability function, we divide the probability of each value of X by the total number of possibilities in the sample space, which is 8.
Therefore, the probability function of X is:
P(X = 0) = 1/8
P(X = 1) = 3/8
P(X = 2) = 3/8
P(X = 3) = 1/8
Note: The total probability adds up to 1, as it should for a probability function.