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13 February, 18:32

There are two coins, one is fair and one is biased. The biased coin has a probability of landing on heads equal to 4/5. One of the coins is chosen at random (50-50), and is flipped repeatedly until it lands on tail. If it landed on heads 4 times before landing on tails, what is the posterior probability that coin chosen was biased?

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  1. 13 February, 18:51
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    72.4%

    Step-by-step explanation:

    The probability of A occurring given that B occurs = the probability of both A and B / the probability of B

    P (A|B) = P (A∩B) / P (B)

    This can be rearranged as:

    P (A∩B) = P (B) P (A|B)

    In this case:

    A = biased coin is chosen

    ~A = fair coin is chosen

    B = 4 heads then 1 tail

    First, let's find P (A∩B).

    P (A∩B) = P (B) P (A|B)

    P (A∩B) = ½ * ₅C₄ (⅘) ⁴ (⅕) ¹

    P (A∩B) = 0.2048

    Next, find P (~A∩B).

    P (~A∩B) = P (B) P (~A|B)

    P (~A∩B) = ½ * ₅C₄ (½) ⁴ (½) ¹

    P (~A∩B) = 0.078125

    Therefore, the probability that the coin is biased is:

    P = P (A∩B) / (P (A∩B) + P (~A∩B))

    P = 0.2048 / (0.2048 + 0.078125)

    P = 0.723866749

    The probability is approximately 72.4%.
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