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:

$\frac{15}{4}$

Step-by-step explanation:

P(2 heads) = P(first flip = head)*P(second flip = head)

=> $(\frac{1}{2} )(\frac{1}{2} )$ = $\frac{1}{4}$

P(1 head) = P(first flip = head)*P(second flip = tail) + P(first flip = tail)*P(second flip = head)

= $\frac{1}{4} +\frac{1}{4} =\frac{1}{2}$

P(no heads) = P(first flip = tail)*P(second flip = tail)

= $\frac{1}{4}$

E(winning) =$10(\frac{1}{4} )+0(\frac{1}{2} )+5 (\frac{1}{4} )$

= $(\frac{10}{4} )+ (\frac{5}{4} )$

= $\frac{15}{4}$