Here is the question of conditional probability:
Consider the experiment of tossing a coin. If the coin shows tails, toss it again but if it shows head, then throw a die. Find the conditional probability of the event that "the die shows a number greater than 3" given that "there is at least one head".
I did it using a TREE diagram as:
{COIN THROW} : i) $\space\space\space\space$ H : {1,2,3,4,5,6}
ii) T : {H,T}
Hence at least one head = {(H,1);(H,2);(H,3);(H,4);(H,5);(H,6);(T,H)}
Hence , P(at least one head) = 7/8
Among these , number greater than 3 : {(H,4);(H,5);(H,6)}
Hence , P = 3/8
Which brings me to P (Required) = $\dfrac{3}{8} * \dfrac{8}{7}$ = 3/7 $\space\space\space\space$[Real ans = 1/3]
Please explain how.
Best Answer
Let $D$ denote the number eventually thrown by the die. Discern $4$ disjoint events:
1) $\left(H\text{ and }D>3\right)$ probability: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
2) $\left(H\text{ and }D\leq3\right)$ probability: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
3) $\left(T\text{ and }H\right)$ probability: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
4) $\left(T\text{ and }T\right)$ probability: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
Then $P\left(D>3\mid\text{at least one head}\right)=\frac{P\left(\text{at least one head and }D>3\right)}{P\left(\text{at least one head}\right)}=\frac{\frac14}{\frac34}=\frac13$
The numerator is the probability event 1)
The denominator is the probability of the union of the events 1),2),3)