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Probability - A coin comes up heads about 60% of the time. If it is tossed 10 times, what is the probability that exactly between 5 and 7 heads occur consecutively?

(I received the following clarification: It is the probability of getting exactly 5 consecutively or exactly 6 consecutively or exactly 7 consecutively.)

I was thinking of a binomial distribution, but not so sure that will address the precise details or is there a chance this is 10 Bernoulli trials and we need to find the conditional probability that all successes will occur consecutively (ie, no two successes will be separated by any failures) and we need to give the number of successes between 5 and 7. Then I think the denominator will be the binomial distribution summing from k=5 to k=7, but then not sure of the numerator precisely.

Please show details so I may understand the correct process.

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  • $\begingroup$ Do you mean that the coin is unfair and every toss will come up heads with $60\%$ probability? The wording of the title suggests something different: we toss a coin whose fairness was not specified, and it comes up heads "about" six times ($60\%$ of $10$). I don't think that's what you mean, particularly in light of the rest of the question, but the title (and the first sentence of the question) are not the clearest possible expression of whatever you do mean. $\endgroup$ – David K Oct 22 '15 at 21:11
  • $\begingroup$ I took the question to be p=0.60. $\endgroup$ – user217189 Oct 22 '15 at 21:50
  • $\begingroup$ I don't think this is clear. Are you asking "if you toss this weighted coin $10$ times what is the probability that you get exactly $6$ Heads?" ? If not, how does your question differ? $\endgroup$ – lulu Oct 22 '15 at 21:55
  • $\begingroup$ The "occur consecutively" makes this problem different than a standard evaluation of the binomial distribution. $\endgroup$ – Paul Oct 22 '15 at 21:58
  • $\begingroup$ Sorry, there has been wording issues on this problem. I am considering it to be p=0.60 and you toss this unfair coin 10 times and need the probability that exactly between 5 and 7 heads occur consecutively, so 5 heads or 6 heads or 7 heads $\endgroup$ – user217189 Oct 22 '15 at 22:34
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You want the probability of getting between five and seven heads in a row.

$P(\text{between 5 and 7 heads in a row}) = P(\text{5 H in a row}) + P(\text{6 H in a row}) + P(\text{7 H in a row})$

I find it best to enumerate (at least initially) the possibilities. For 7 heads, you need:

(1) $HHHHHHHTxx$ or

(2) $THHHHHHHTx$ or

(3) $xTHHHHHHHT$ or

(4) $xxTHHHHHHH$

where the $x$ doesn't matter because it doesn't affect our final result.

What is the probability of getting (1)? That can be done easily, where (1) is now $0.6^7\times0.4$. Similarly, the probability of (2) is $0.4^2\times0.6^7$. Note that (3) is equivalent to (2) and (4) is equivalent to (1). Thus, $P(\text{7 H in a row})=2\times0.6^7\times0.4+2\times0.4^2\times0.6^7$.

You can do the same thing with 6 heads and 5 heads. There will be a few more cases than four, but it shouldn't be computationally prohibitive. Recognizing that cases will "double up" (i.e. (2)/(3) and (1)/(4) above) will simplify calculations tremendously.

There is perhaps an easier and more sophisticated way of doing this, but this is how I would attack this problem intuitively. You could possibly find a generalization rule by collecting the 0.4 and 0.6 terms at the end.

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  • $\begingroup$ I received the following clarification: It is the probability of getting exactly 5 consecutively or exactly 6 consecutively or exactly 7 consecutively. $\endgroup$ – user217189 Oct 22 '15 at 23:07
  • $\begingroup$ So the above should hold, as the $T$ in (1)-(4) indicate that it is exactly (in this case) seven heads in a row. You need to make sure that you have a $T$ preceding and succeeding the set of heads when they do not fall on the end. $\endgroup$ – Matt Brems Oct 22 '15 at 23:10
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    $\begingroup$ Unless you're saying that there are exactly five heads of ten coins and they all happen consecutively. I interpreted this to mean that in a sequence of ten coins, there exists some sequence of exactly five heads. $\endgroup$ – Matt Brems Oct 22 '15 at 23:11
  • $\begingroup$ Less ambiguous wording for the question would be "exactly 5, all consecutive, exactly 6, all consecutive, or exactly 7, all consecutive" (leading to the answer by rtybase) or "a run of at least 5 and no more than 7 heads", leading to this answer. $\endgroup$ – David K Oct 23 '15 at 0:24
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    $\begingroup$ User, that implies that no heads can occur outside of the run we see. That does not seem to be the interpretation of the question. I believe more heads can occur, but only 5-7 in a row. I think the question was worded as such to imply that a run of ten heads would not count as a valid option. $\endgroup$ – Matt Brems Oct 23 '15 at 0:40
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Strictly "between 5 and 7" would mean it comes up heads 6 times out of 10. Given that the probability of heads on any one toss, that would be the binomial probability $\begin{pmatrix}10 \\ 6\end{pmatrix}0.6^60.4^4$. If you mean "5 or 6 or 7" the probability is $\begin{pmatrix}10 \\ 5\end{pmatrix}0.6^50.4^5+ \begin{pmatrix}10 \\ 6 \end{pmatrix}0.6^60.4^4+ \begin{pmatrix}10 \\ 7\end{pmatrix}0.6^70.4^3$.

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    $\begingroup$ But these heads are not necessarily consecutive. $\endgroup$ – Paul Oct 22 '15 at 22:06
  • $\begingroup$ "occur consecutively" - you missed this part. So you can't use combinations ... $\endgroup$ – rtybase Oct 22 '15 at 22:06
  • $\begingroup$ Is there a chance this is 10 Bernoulli trials and we need to find the conditional probability that all successes will occur consecutively (ie, no two successes will be separated by any failures) and we need to give the number of successes between 5 and 7. Then I think the denominator will be the binomial distribution summing from k=5 to k=7, but then not sure of the numerator precisely. $\endgroup$ – user217189 Oct 22 '15 at 22:45
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Golden rule with discrete probabilities, if there is no formula ready to use, model and compute the outcome:

5:

  • H-H-H-H-H-T-T-T-T-T
  • T-H-H-H-H-H-T-T-T-T
  • T-T-H-H-H-H-H-T-T-T
  • T-T-T-H-H-H-H-H-T-T
  • T-T-T-T-H-H-H-H-H-T
  • T-T-T-T-T-H-H-H-H-H $$6\cdot 0.6^5\cdot0.4^5$$

Or 6:

  • H-H-H-H-H-H-T-T-T-T
  • T-H-H-H-H-H-H-T-T-T
  • T-T-H-H-H-H-H-H-T-T
  • T-T-T-H-H-H-H-H-H-T
  • T-T-T-T-H-H-H-H-H-H $$5\cdot 0.6^6\cdot0.4^4$$

Or 7:

  • H-H-H-H-H-H-H-T-T-T
  • T-H-H-H-H-H-H-H-T-T
  • T-T-H-H-H-H-H-H-H-T
  • T-T-T-H-H-H-H-H-H-H $$4\cdot 0.6^7\cdot0.4^3$$ Should be easy to compute by hand.
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  • $\begingroup$ But the question does not require that all of the heads appear consecutively; it just requires that 5, 6, or 7 do. So for example HHHHHTHHHH is another possibility. $\endgroup$ – Paul Oct 22 '15 at 22:26
  • $\begingroup$ It does ... there is an explanation in the comments "and you toss this unfair coin 10 times and need the probability that exactly between 5 and 7 heads occur consecutively, so 5 heads or 6 heads or 7 heads" $\endgroup$ – rtybase Oct 22 '15 at 22:28
  • $\begingroup$ It's that mixture of "exactly" and "consecutive" ... might sound confusing. At least I highlighted a way to attack the problem. $\endgroup$ – rtybase Oct 22 '15 at 23:00
  • $\begingroup$ what are you thinking? $\endgroup$ – user217189 Oct 22 '15 at 23:02
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    $\begingroup$ Well, if it's about "(exactly 5 and all consecutive) or (exactly 6 and all consecutive) or (exactly 7 and all consecutive)" then you have the answer. Otherwise, see the answer posted by Matt Brems. Essentially, the idea is similar. $\endgroup$ – rtybase Oct 22 '15 at 23:07
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$\begin{pmatrix}10 \\ 6\end{pmatrix}0.6^60.4^4=210\cdot 0.6^60.4^4=.25082$

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