Probability - A coin is tossed 10 times and comes up heads about 60% of the time. What is ... 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.
 A: 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.
A: 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 $\begin{pmatrix}10 \\ 5\end{pmatrix}0.6^50.4^5+ \begin{pmatrix}10 \\ 6 \end{pmatrix}0.6^60.4^4+ \begin{pmatrix}10 \\ 7\end{pmatrix}0.6^70.4^3$.
A: 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.

A: $\begin{pmatrix}10 \\ 6\end{pmatrix}0.6^60.4^4=210\cdot 0.6^60.4^4=.25082$
