Question

Determine $P\left(\frac{E}{F}\right)$
Mother, father and son line up at random for a family picture

Answer 1

If mother (m) , father (i) and son (s) line for the family picture, then the sample space will be S = {mfs, msf, fms, fsm, smf, sfm}
$\Rightarrow n\left(S\right)=6$ Which contains 6 equally likely sample events.
here, E = set of events having son on one end and
F= set of event having father in the middle
$E=mfs,fms,smf,sfm,F=mfs,sfmandE\cap F=mfs,sfm\Rightarrow n\left(E\right)=4,n\left(F\right)=2,m)E\cap F)=2.Now,P\left(E\right)=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{4}{6}=\frac{2}{3}$
Similarly, $P\left(F\right)=\frac{2}{6}=\frac{1}{3}andP(E\cap F)=\frac{2}{6}=\frac{1}{3}$
We know that P $\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$