A coin is tossed till Head appears or until three tosses have been completed(whichever is earlier). Find probability that coin is tossed thrice given that first toss is not Heads.

My try:

Given first toss is not heads it implies first toss is tails. So for three tosses we should get $TTH$ Or $TTT$ whose probability is $\frac{1}{8}\times 2=0.25$

But answer is $\frac{3}{8}$. What is wrong in my solution?

  • 3
    $\begingroup$ I don't agree with either answer. The first toss is given as $T$, that's not a matter of probability. Thus the only issue is whether the second toss is $T$ so the answer is $\frac 12$. $\endgroup$
    – lulu
    Nov 3, 2017 at 17:40

1 Answer 1


It is worth going through the effort of calculating the probability via the definition of conditional probability in early examples.

$$Pr(A\mid B):=\frac{Pr(A\cap B)}{Pr(B)}$$

Let $B$ be the event that the first coin flipped is not a head (i.e. the first coin flipped turned up tails).

Let $A$ be the event that the coin is flipped exactly three times.

We are tasked with calculating $Pr(A\mid B)$, the probability that the coin is flipped exactly three times given that the first flip did not turn up heads.

We can draw ourselves a tree diagram or however else we like to arrive at the following table of outcomes and respective probabilities:

$$\begin{array}{|c|c|}\hline\text{Outcome}&\text{Probability}\\\hline H&\frac{1}{2}\\\hline TH&\frac{1}{4}\\\hline TTH&\frac{1}{8}\\\hline TTT&\frac{1}{8}\\\hline\end{array}$$

It is worth taking a moment to check that this does in fact make sense as a probability distribution by verifying that the probabilities add up to exactly one. Indeed $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}$ does equal $1$.

The event that the first flip is not heads corresponds to all of the above listed outcomes except the first and so occurs with probability $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$ so we learn that $Pr(B)=\frac{1}{2}$.

The event that the first flip is not heads and it takes three flips in total corresponds to the last two outcomes in the above table and so occurs with probability $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ so we learn that $Pr(A\cap B)=\frac{1}{4}$.

Putting this information together, we get:

$$Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{1/4}{1/2}=\frac{1}{2}$$


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