Math

posted by Steve

A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times
a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?
b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.
here is what I tried

a) p(at least 2 h | 1 h) = p(at least 1h and at least 2 h)/p(at least one head 1h)
I get stuck here as to what numbers I put I know that p(no heads in toss 1)= 1- 2/3 = 1/3

I appreciate any help
Thanks

  1. bobpursley

    https://www.physicsforums.com/threads/probability-unfair-coin-toss-probably-pretty-easy.342119/

  2. Steve

    I looked at this link and still cant get it

  3. Reiny

    Lets' do the the prob of each of the two events, then worry about the conditional probability.

    Let prob(A) be the prob(at least one head in the three tosses)
    = 1 - prob(no head three times)
    prob(head) = 2/3, prob(no head) = 1/3
    prob(A) = 1 - C(3,0)(1/3)^3 (2/3)^0 = 1/27 = 26/27

    let prob(B) be prob(at least two heads in 3tries)
    = prob(exactly 2) + prob(exactly 3)
    = C(3,2)((2/3)^2 (1/3) + C(3,3)(2/3)^3 (1/3)^0
    = 3(4/9)(1/3) + 1(8/27)(1)
    = 12/27 + 8/27 = 20/27

    so you have P(P|Q) = prob( P and Q)/prob(P)
    read as : prob( P given Q)
    carry on

    or

    There are only 8 outcomes, so lets list them and their probs
    HHH ----> (2/3)^3 = 8/27
    HHT ----> (2/3^2 (1/3) = 4/27
    HTH ----> (2/3)^2 (1/3) = 4/27
    THH ----> (2/3)^2 (1/3) = 4/27
    HTT ----> (2/3)(1/3)^2 = 2/27
    THT ----> (2/3)(1/3)^2 = 2/27
    TTH ----> (2/3)(1/3)^2 = 2/27
    TTT ----> (1/3)^3 = 1/27 , note that their sum is 1 , as needed.

    out of those with at least 1 head, which is a prob of 26/27, there are 4 cases of at least 2 H's , the sum of those = 8/27 + 3(4/27) = 20/27
    so prob(what you asked for) = (20/27) / (26/27) = 20/26 = 10/13

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