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Math Help - Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

  1. #1
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    Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

    Hey, I have the following question... I know the probabilities for each event happening independently, but not in the way the question is asking. Should I multiply or add the probabilities? What should I do? Cheers!

    The problem says that person A tosses a fair coin until it lands heads, and that then person B does the same. In the end, they toss it 10 times. What is the probability that person A only tossed it once?

    Well, I have that the probability that person A gets heads with only once throw is 0.5, and that the probability that person gets heads with 9 throws is 9/(2^9)=9/512=0.02

    So I don't know if I should add them (which would give P=0.52) or multiply them... or something different. Thanks a lot!
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  2. #2
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    Re: Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

    Hello, buenogilabert!

    Person A tosses a fair coin until it lands heads, then person B does the same.
    In the end, they toss it 10 times.
    What is the probability that person A tossed it only once?

    A tossed once and got a Head,
    . . then B tossed nine times to grt a Head.

    We want: . P(\text{A: Head on 1st toss}) \;\text{ and }\;P(\text{B: 8 Tails, then a Head})

    . . . . . . . . . . . . . . . . \frac{1}{2} \qquad\qquad\quad\; \times \qquad\qquad\quad \left(\frac{1}{2}\right)^8\cdot\frac{1}{2}


    Therefore: . \text{Probability} \:=\:\left(\frac{1}{2}\right)^{10} \:=\:\frac{1}{1024}

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