Suppose an experiment consists of tossing a fair coin until three heads occur. What is the probability that experiment ends after exactly six flips of the coin with a head on the fifth toss as well as sixth?

Now I know how to solve for the case; when I have to find probability of three heads to occur in 6 independent tosses of fair coin. I use the binomial distribution with $n$=6 , $x$=3 and get result $P(x=3)$ = 5/16
What confuses me is the last condition of getting head on fifth and sixth trial. How do I proceed?

  • $\begingroup$ The condition implies that in the first $4$ trials, exactly one "heads" must occur. In the title, it is assumed that the fifth ans sixth trial is "heads", in the body however not. Please clarify which is the intented situation. $\endgroup$ – Peter Jan 3 at 11:32
  • $\begingroup$ It is clarified at the end of first paragraph in the body, that head occurs in the 5th as well as 6th trial. $\endgroup$ – Ameer Shah Jan 3 at 11:35
  • $\begingroup$ My question is whether this is given. In this case, we have a conditional probability. Or whether it is required that they are "heads" but not given. In this case, you have to mutlitply the probability of exactly one "heads" in $4$ trials with $\frac{1}{4}$ $\endgroup$ – Peter Jan 3 at 11:38
  • $\begingroup$ I think It is required. I have wrote the exact question word by word written in the book, in the first paragraph of the body of question. What do you think? $\endgroup$ – Ameer Shah Jan 3 at 11:44
  • $\begingroup$ I think it is only required, so you have to multiply with $\frac{1}{4}$ $\endgroup$ – Peter Jan 3 at 12:18


To get the experiment to end with three heads after exactly six flips of the coin with a head on the fifth toss as well as sixth, you need

  • exactly one of the first four to be heads
  • the fifth to be heads
  • the sixth to be heads
  • $\begingroup$ This suggests that I use binomial distribution for the first case and negative binomial distribution for the last two cases (when head occurs on fifth as well as sixth toss), then add the two results. Am I right? $\endgroup$ – Ameer Shah Jan 3 at 11:39
  • $\begingroup$ @AmeerShah You can treat each of the last two events as Bernoulli (or binomial with $n=1$) distributed. Obviously they each occur with probability $\frac12$ as you have a fair coin $\endgroup$ – Henry Jan 3 at 11:42
  • $\begingroup$ I got the answer 1/16 but by multiplying all three results from three cases. (i.e: 1/4*1/2*1/2 = 1/16). I don't get the logic that why are we multiplying but not adding the results? $\endgroup$ – Ameer Shah Jan 3 at 12:07
  • $\begingroup$ @AmeerShah The probabilities of independent events multiply. $\endgroup$ – Henry Jan 3 at 12:33

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