Question

Determine $P\left(\frac{E}{F}\right)$
A die is thrown three times
E: 4 appears on the third toss
F: 6 and 5 appears, respectively on first two tosses.

Answer 1

If a die is thrown three times , then the number of elements in the sample space is S = $\left(\right(x,y,z):x,y,z,ϵ(1,2,3,4.5,6)$
Here, E : set of events in which 4 appears on the third toss and F : set of events in which 6 and 5 appears respectively on first two tosses
$$\left\{\begin{array}{cccccc}(1,1,4),& (1,2,4),& (1,3,4),& (1,4,4),& (1,5,4),& (1,6,4)\\ (2,1,4),& (2,2,4),& (2,3,4),& (2,4,4),& (2,5,4),& (2,6,4)\\ (3,1,4),& (3,2,4),& (3,3,4),& (3,4,4),& (3,5,4),& (3,6,4)\\ (4,1,4),& (4,2,4),& (4,3,4),& (4,4,4),& (4,5,4),& (4,6,4)\\ (5,1,4),& (5,2,4),& (5,3,4),& (5,4,4),& (5,5,4),& (5,6,4)\\ (6,1,4),& (6,2,4),& (6,3,4),& (6,4,4),& (6,5,4),& (6,6,4)\end{array}\right\}$$
And $F=(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,5),(6,5,6)\Rightarrow n\left(F\right)=6\therefore P\left(E\right)=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{36}{216}=\frac{1}{6}Similarly,P\left(F\right)=\frac{6}{216}=\frac{1}{36}andP(E\cap F)=\frac{1}{216}Requiredprobability,P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{216}}{\frac{1}{36}}=\frac{1}{216}\times \frac{36}{1}=\frac{1}{6}$