Suppose you have a bag of 10 coins. Nine of them are fair coins, that is, if you toss any of these 9 coins the probability of getting a head

Question

Suppose you have a bag of 10 coins. Nine of them are fair coins, that is, if you toss any of these 9 coins the probability of getting a head, P(H) = 1/2. Similarly, probability of getting a tail, P(T) = 1/2. The other coin is biased — it has head on both sides. Use indicator random variables to compute expected number of heads, if all 10 coins are tossed together.

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Raelynn 1 week 2021-10-08T00:52:50+00:00 1 Answer 0

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    2021-10-08T00:54:26+00:00

    Answer:

    Step-by-step explanation:

    Given:

    we have 10 coins

    nine of them are fair coins, which means: P(H)=\frac{1}{2}; P(T)=\frac{1}{2}

    but one coin is biased it has head on both sides which means P(H)=\frac{1}{1}; P(T)=\frac{0}{1}=0

    expected number of heads for tossing 9 coins = 9(\frac{1}{2})=\frac{9}{2}=4.5

    The coin 10 is biased  with only head on both sides

    The expected number of heads tossing this coin is = 1

    Therefore, the expected number of heads if all 10 coins are tossed together is =4.5+1=5.5

    Using indicator random variables:

    The number of unbiased coins = 9: n=9

    P(H)=\frac{1}{2}=0.5\\\\F(x)=nP=9\times 0.5=9.5

    One coin is biased with only head. Therefore:

    F(x)=1

    Finally, F(x)=4.5+1=5.5

    The expected number of heads = 5.5

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