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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?

Respuesta :

Answer:

[tex]\frac{15}{4}[/tex]

Step-by-step explanation:

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

=> [tex](\frac{1}{2} )(\frac{1}{2} )[/tex] = [tex]\frac{1}{4}[/tex]

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

= [tex]\frac{1}{4} +\frac{1}{4} =\frac{1}{2}[/tex]

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

= [tex]\frac{1}{4}[/tex]

E(winning) =[tex]10(\frac{1}{4} )+0(\frac{1}{2} )+5 (\frac{1}{4} )[/tex]

= [tex](\frac{10}{4} )+ (\frac{5}{4} )[/tex]

= [tex]\frac{15}{4}[/tex]