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A coin is tossed twice. If heads comes up either time, we get $2. If heads does not come up, we lose $4. What is the expected value of this game?
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probability of getting at least one head on two tosses is \[\frac{3}{4}\] and of getting no heads is \[\frac{1}{4}\]
\[E=2\times \frac{3}{4}-4\times \frac{1}{4}=\frac{6}{4}-\frac{4}{4}=\frac{2}{4}=\frac{1}{2}\]
or 50 cents. unless the problem meant you get $2 for each head that comes up, but i did not read it that way.
thanks
welcome. E= what you get times probability you get it.
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