Question

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

• A. $\frac{7}{9}$
• B. $\frac{2}{7}$
• C. $\frac{2}{9}$
• D. $\frac{4}{9}$

Answer 1

The correct option is C $\frac{2}{9}$

The outcomes of the experiment can be represented in following diagramatic manner called the 'tree diagram'.
The sample space of the experiement may be described as
$S=\left\{\right(H,H),(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6\left)\right\}$
Where $(H,H)$ denotes that both the tosses result into head and $(T,i)$ denotes 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 the 8 elementary events
$(H,H),(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)$ 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 which is clear from the given figure

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
$F=\left\{\right(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6\left)\right\}$
$E=\left\{\right(T,5),(T,6\left)\right\}$
and $E\cap F=\left\{\right(T,5),(T,6\left)\right\}$
Now $P\left(F\right)=P\left(\left\{\right(H,T\left)\right\}\right)+P\left(\left\{\right(T,1\left)\right\}\right)+P\left(\left\{\right(T,2\left)\right\}\right)+P\left(\left\{\right(T,3\left)\right\}\right)+P\left(\left\{\right(T,14\left)\right\}\right)+P\left(\left\{\right(T,5\left)\right\}\right)+P\left(\left\{\right(T,6\left)\right\}\right)$
$=\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{4}$
$P(E\cap F)=P\left(\left\{\right(T,5\left)\right\}\right)+P\left(\left\{\right(T,6\left)\right\}\right)=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$
Hence $P\left(E/F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{6}}{\frac{3}{4}}=\frac{2}{9}$