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ijohnh14
14.12.2019 •
Mathematics
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?
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Ответ:
Step-by-step explanation:
P(2 heads) = P(first flip = head)*P(second flip = head)
=>
= ![\frac{1}{4}](/tpl/images/0417/8728/0d86b.png)
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}](/tpl/images/0417/8728/caad2.png)
P(no heads) = P(first flip = tail)*P(second flip = tail)
=![\frac{1}{4}](/tpl/images/0417/8728/0d86b.png)
E(winning) =![10(\frac{1}{4} )+0(\frac{1}{2} )+5 (\frac{1}{4} )](/tpl/images/0417/8728/a62eb.png)
=![(\frac{10}{4} )+ (\frac{5}{4} )](/tpl/images/0417/8728/c7752.png)
=![\frac{15}{4}](/tpl/images/0417/8728/c42ae.png)
Ответ:
niger
Step-by-step explanation:
niger