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.

  • $\begingroup$ How many times did you toss the coin? Do re-tosses count as tosses? $\endgroup$ – snar Feb 21 '15 at 19:16
  • $\begingroup$ The probability that twice you toss a tail is $\frac14$. The probability that you toss a tail and then a head is $\frac14$ (not $\frac18$). Notice e.g. that $(T,H)$ is more likely than $(H,1)$. You treat the probabilities as if they are the same. $\endgroup$ – drhab Feb 21 '15 at 19:16

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)


outcome $(H,4)$ has probability $(1/2) \times (1/6)$

outcome $(H,5)$ has probability $(1/2) \times (1/6)$

outcome $(H,6)$ has probability $(1/2) \times (1/6)$

Therefore the probability of a favourable result is $1/4$

That at least one head has probability $(1/2)+(1/4)=3/4$

Thus on your contingency tree the favourable leaves have probability $1/4$ and the probability of leaves with at least one head (which of course include all the favourable leaves) have probability $3/4$. Hence the conditional probability is $(1/4)/(3/4)=1/3$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.