Prob(2 heads | first flip is head)

[operationsres]

EDIT: I think I may have solved it a few minutes after I posted. See below for proposed solution ...

Homework Statement

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

2. The attempt at a solution

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

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

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

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EDIT: I think I might have solved it. Let A = {HH,HT} and B = {HH}, then A$\cap$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 ...

 

[uart]

EDIT: I think I might have solved it. Let A = {HH,HT} and B = {HH}, then A$\cap$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.

Yes that's correct. :)