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Prob(2 heads | first flip is head)

  1. Jul 31, 2012 #1
    EDIT: I think I may have solved it a few minutes after I posted. See below for proposed solution ...

    1. The problem statement, all variables and given/known data

    Determine [itex]\textrm{Prob}(2 \textrm{heads}|\textrm{first flip is head})[/itex] by using the formula [itex]P(B|A) = \frac{P(B \cap A)}{P(A)}[/itex]. Specifically, determine what the sets A and B are .

    2. The attempt at a solution

    Clearly the state space has collapsed to [itex]\Omega = \{HH, HT\}[/itex], and the [itex]\sigma[/itex]-algebra is [itex]\bf{F}=\{\{ \},\Omega,\{HH\},\{HT\}\}[/itex]. Let Z be the event [itex]Z = \{HH\}[/itex] within [itex]\Omega[/itex].

    The probability is easily computed as the cardinality of Z divided by the cardinality of the [itex]\Omega[/itex], i.e. Prob(2 heads|first flip is head) = 0.5.

    _____________________

    My problem is that I can't figure out what sets A and B are supposed to be such that [itex]P(B|A) = \frac{P(B \cap A)}{P(A)}[/itex] gives me 50%.

    _____________________

    EDIT: I think I might have solved it. Let A = {HH,HT} and B = {HH}, then A[itex]\cap[/itex]B = {HH}, and Prob({HH})=0.25.

    The denominator is P({HH,HT}) = 2/4 = 0.5.

    0.25/0.5 = 50%, which is the correct solution.


    Feel free to delete if this is correct ...
     
  2. jcsd
  3. Jul 31, 2012 #2

    uart

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    Science Advisor

    Re: [itex]\textrm{Prob}(2 \textrm{heads}|\textrm{first flip is head})[/itex]

    Yes that's correct. :)
     
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