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Coin toss probability problem

  1. Dec 16, 2009 #1
    Hi all =)

    Question...
    A fair coin is tossed thrice. Supposed we denote a "head" turning up as 1 and
    "tail" as 0. Given that the total on all three tosses is an odd number,
    what is the probability that at first toss, we get a "head"?

    i dont have any idea to start answering this question..
    please guide me.. =)
     
  2. jcsd
  3. Dec 16, 2009 #2
    Re: Probability

    Ok our sample space is
    [tex]\Omega = \{0,1\}^3 = \{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}[/tex]
    We assume a fair coin so every outcome is equally likely which gives us the probability function:
    [tex]p(x) = \frac{1}{8} \qquad \textrm{for all }x \in \Omega[/tex]
    Now we have two events [itex]E_1,E_2[/itex]. Let [itex]E_2[/itex] denote the event that the total on all three tosses is an odd number. Let [itex]E_1[/itex] denote the event that on the first toss we get a head. What you want is the conditional probability [itex]P(E_1|E_2)[/itex]. List all elements in [itex]E_1[/itex] and [itex]E_2[/itex] and you should be able to compute it using the formula:
    [tex]P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}[/tex]
     
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