Question

Determine $P\left(\frac{E}{F}\right)$
A coin is tossed three times
E: head on third toss F: head on first two tosses

E: Atleast two heads F : atmost two heads

E: Atmost two tails F : atleast one tail

Answer 1

When a coin is tossed is three times, the sample space S Contains 2^3 = 8 equally likely sample points.
$\because$ S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Here, E : set of events in which head occurs on third toss and : set of events in whichhead occurs on third toss and F : set of events in which heads occur on first two tosses,
$\therefore E=HHH,THH,HTH,TTHandF=(HHH,HHT)\Rightarrow E\cap F=\left(HHH\right)$
Hence, $P\left(E\right)=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes}=\frac{4}{8}=\frac{1}{2}Similarly,P\left(F\right)=\frac{2}{8}=\frac{1}{4}andP(E\cap F)=\frac{1}{8}$
$\therefore$ From the formula $P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{8}\times \frac{1}{4}=\frac{1}{2}$

Here, E : set of events in which atleast two heads occur and F: set of events in which atmost tow heads occur.
$\therefore E=(HHH,HHT,HTH,THH)$
[Here, we consider the cases of two and three heads, as the condition given is atleast not atmost]
F = (TTT, THT, TTH, HTT, HHT, HTH, THH,)
$\Rightarrow (E\cap F)=(HHT,HTH,THH)$
Hence $P\left(E\right)=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{4}{8}=\frac{1}{2}SimilarlyP\left(F\right)=\frac{7}{8}andP(E\cap F)=\frac{3}{8}\therefore P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{8}\times \frac{8}{7}=\frac{3}{7}$

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) = $\frac{7}{8}andP(E\cap F)=\frac{6}{8}\therefore \left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$