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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    probability of getting at least one head on two tosses is \[\frac{3}{4}\] and of getting no heads is \[\frac{1}{4}\]

  2. anonymous
    • 5 years ago
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    \[E=2\times \frac{3}{4}-4\times \frac{1}{4}=\frac{6}{4}-\frac{4}{4}=\frac{2}{4}=\frac{1}{2}\]

  3. anonymous
    • 5 years ago
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    or 50 cents. unless the problem meant you get $2 for each head that comes up, but i did not read it that way.

  4. anonymous
    • 5 years ago
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    thanks

  5. anonymous
    • 5 years ago
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    welcome. E= what you get times probability you get it.

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