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 of the event 'the coin shows a tail', given that 'at least one die shows a $3$'.

Answer 1

The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
$S=\{(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Let $A$ be the event that the coin shows a tail and $B$ be the event that at least one die shows $3$.
$\therefore A=\{(1,T),(2,T),(4,T),(5,T)\}$
$B=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}$
$\Rightarrow A\cap B=\varphi$
$\therefore P(A\cap B)=0$
Then, $P\left(B\right)=P\left(\{3,1\}\right)+P\left(\{3,2\}\right)+P\left(\{3,3\}\right)+P\left(\{3,4\}\right)+P\left(\{3,5\}\right)+P\left(\{3,6\}\right)+P\left(\{6,3\}\right)$
$=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}$
$=\frac{7}{36}$
Probability of the event that the coin shows a tail, given that at least one die shows $3$, is given by $P\left(A|B\right)$.
Therefore,$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{0}{\frac{7}{36}}=0$