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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.

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  • $\begingroup$ How many times did you toss the coin? Do re-tosses count as tosses? $\endgroup$
    – snar
    Commented Feb 21, 2015 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
    Commented Feb 21, 2015 at 19:16

2 Answers 2

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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)

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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 $(1/4)/(3/4)=1/3$

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