Question

In a game you flip a coin twice, and
record the number of heads that occur.
You get 10 points for 2 heads, zero
points for 1 head, and 5 points for no
heads. What is the expected value for
the number of points you’ll win per
turn?

Answer 1

Answer on Question#38988 - Math - Statistics and Probability

Question: In a game you flip a coin twice, and record the number of heads that occur. You get 10 points for 2 heads, zero points for 1 head, and 5 points for no heads. What is the expected value for the number of points you'll win per turn?

Solution:

P(2 heads) = P(first flip = head)*P(second flip = head) = (1/2)(1/2) = 1/4.

P(1 head) = P(first flip = head)*P(second flip = tail) + P(first flip = tail)*P(second flip = head) = 1/4 + 1/4 = 1/2.

P(no heads) = P(first flip = tail)*P(second flip = tail) = 1/4.

E(winning) = 10*P(winning = 10) + 0*P(winning = 0) + 5*P(winning = 5) = 10*(1/4) + 0*(1/2) + 5*(1/4) = 15/4.

Answer: 15/4.