A coin is tossed 7 times. The probability that at least 4
consecutive heads appear is?
I have checked and double checked but I can't figure out which cases I am missing.
Here are the cases:
- 4 consecutive heads appear:
$n_1 = \dfrac{4!}{3!}= 4$
- 5 heads appear
Subcase 1: 4 consecutive heads appear and the other head is separated.
For example: HTTHHHH or THTHHHH
Subcase 1 has $4$ possibilities.
Subcase 2: 5 consecutive heads:
$\dfrac{3!}{2!} = 3$
$n_2 = 4+ 3 = 7$
- 6 heads appear
There are 6 gaps (including the ends) where we can place the $T$
Let me show those places by G:
GHGHGHHGHGHG
$n_3 = 6$
- All heads appear
$n_4 = 1$
$P(\text{4 consecutive heads}) = \dfrac{n_1+ n_2+ n_3 + n_4}{2^7} = \dfrac{18}{2^7} = \dfrac{9}{64}$
Please let me know my mistake.
Best Answer
The component of $n_2$ counting the ways that five heads appear, one of which is separated from the other four, should be 6 rather than 4: $$\text{HTTHHHH THTHHHH}\\ \text{HTHHHHT THHHHTH}\\ \text{HHHHTTH HHHHTHT}$$ Thus the probability is $\frac{20}{2^7}=\frac5{32}$.