Answer
3/7.
Work Step by Step
There are ${{2}^{3}}=8$ possible outcomes after tossing a fair coin fairly 3 times. The sample space is:
$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$
Let A be the event that at least two heads will occur and B be the even that at most two heads have occurred.
We see that three of the sample outcomes in S constitute the event A:
$A=\{HHH,HHT,HTH,THH\}$
\[P(A)=\frac{4}{8}\]
We see that seven of the sample outcomes in S constitute the event B:
$B=\{HHT,HTH,HTT,THH,THT,TTH,TTT\}$
\[P(B)=\frac{7}{8}\]
$A\cap B=\{HHT,HTH,THH\}$
\[P(A\cap B)=\frac{3}{8}\]
We have to find the probability of that at least two heads will occur given that at most two heads have occurred:
\[\begin{align}
& P(A|B)=\frac{P(A\cap B)}{P(B)} \\
& P(A|B)=\frac{{}^{3}/{}_{8}}{{}^{7}/{}_{8}} \\
& P(A|B)=\frac{3}{7} \\
\end{align}\]
