Question

Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'.

Answer 1

The sample space of experiment may be described as
$S=(H,H),(H,T),(T,1),(T,2)(T,3)(T,4)(T,5)(T,6)$
where $(H,H)$ denotes that both the tosses result into head and $(T,i)$ denote the first toss result into a tail and the number 'i' appeared on the die for $i=1,2,3,4,5,6$.
Thus, the probabilities assigned to 8 elementary events are $\frac{1}{4},\frac{1}{4},\frac{1}{12},\frac{1}{12},\frac{1}{12},\frac{1}{12},\frac{1}{12},\frac{1}{12}$ respectively.
Let F be the event that 'there is atleast one tail' and E be the event 'the die shows a number greater than 4'. Then,
$F=(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)$
$E=(T,5),(T,6)$ and $E\cap F=(T,5),(T,6)$
Now, $P\left(F\right)=P\left((H,T)\right)+P\left((T,1)\right)+P\left((T,2)\right)+P\left((T,3)\right)+P\left((T,4)\right)+P\left((T,5)\right)+P\left((T,6)\right)$
$=\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{4}$
and $P(E\cap F)=P\left(\left\{(T,5)\right\}\right)+P\left(\left\{(T,6)\right\}\right)=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$
Hence, $P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{6}}{\frac{3}{4}}=\frac{2}{9}$