Question

Determine $P\left(\frac{E}{F}\right)$
A coin is tossed three times
E: Atmost two tails F : atleast one tail

Answer 1

Here E: set of events in which atmost two tails occur and F: set of events in which atleast one tail occurs.
$\therefore$ E = {HHH, HHT, HTH, THH, HTT, THT, TTH}
(here we can consider the cases of the two tails, one tail and no tail because the condition is atmost not atleast)
And F = {TTT, THT, TTH, HTT, HHT, HTH, THH)
(Here, we can consider the cases of one or more tails because the condition given is atleast not atmost)
$\Rightarrow (E\cap F)=(HHT,HTH,THH,HTT,THT,TTH)Now,P\left(E\right)=\frac{NUmberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{7}{8}$
Similarly, P(F) = $\frac{7}{8}andP(E\cap F)$ = $\frac{6}{8}$ $\therefore \left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$