1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

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

    Yes that's correct. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook