Question

Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes, toss a coin. Find the conditional probability fo the event the coin shows a tail, given that atleast one die shows a 3.

Answer 1

The outcomes of the given experiment can be represented by the following set.
The sample space of the experiment is
S= $$\left\{\begin{array}{cccccc}(3,1),& (3,2),& (3,3),& (3,4),& (3,5),& (3,6)\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5),& (6,6)\\ (1,H),& (1,T),& (2,H),& (2,T),& (4,H)& \\ (4,T),& (5,H),& (5,T)& & & \end{array}\right\}$$
n(S) = 20
Let E: the coin shows a tail, F: atleast one die shows up a 3,
$E=\left\{\right(1,T),(2,T),(4,T),(5,T),\Rightarrow n(E)=4$
$F=\left\{\right(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3\left)\right\}\Rightarrow n\left(F\right)=7\Rightarrow E\cap F=\varphi$ because here is no common elements.
$P\left(E\right)=\frac{Favourabeoutcomes}{Totalnumberofoutcomes}\frac{n\left(E\right)}{n\left(s\right)}=\frac{4}{20}=\frac{1}{5}Similarly,P\left(F\right)=\frac{n\left(F\right)}{n\left(S\right)}=\frac{7}{20}andP(E\cap F)=\frac{n(E\cap C)}{n\left(S\right)}=\frac{0}{20}=0$
Hence, the required probability $=P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{0}{\frac{7}{20}}=0$