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4 November, 10:42

You plan on simultaneously throwing seven fair coins. If you get at least twice as many headsas tails, you'll stop; otherwise, you'll throw all seven coins again. How many throws do you expectit will take to stop? [WISE]

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  1. 4 November, 13:52
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    the expected value of throws is E (Y) = 15 shots

    Step-by-step explanation:

    We will divide the experiment in 2

    first) throw seven fair coins, and you get at least twice as many heads as tails when the number of heads is 6 or 7

    being the random variable X=getting x heads, then P (X) has a binomial distribution, with p=0.5 and n=7. then

    p₂=P (X≥6) = P (X=6) + P (X=7) = 0.0546 + 0.0078 = 0.0624

    second) if you get at least twice as many heads as tails when the number of heads, you throw again. then the random variable Y = number of shots in the experiment, then Y follows a negative binomial distribution, with probability of success pn=1-p₂, and number of failures until the experiment is stopped r=1

    then the expected value of throws E (Y) is

    E (Y) = pn*r / (1-pn) = (1-p₂) * 1/p₂ = 1/p₂ - 1 = 1/0.0624 - 1 = 15

    E (Y) = 15 shots
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