Need help!

ActuaryKen5 · #1 12-10-2017, 10:50 PM

An unbiased coin is tossed repeatedly. What is the probability that the fourth occurrence of a head will take place before the second occurrence of a tail?

I don't understand why the answer is 3/16 instead of 5/32.
Can someone help me out on this?
Much appreciated!

Academic Actuary · #2 12-10-2017, 11:24 PM

Four heads on the first four flips, or three heads on the first four with a head on the fifth.

Z3ta · #3 12-11-2017, 01:28 AM

Let $X$ be the number of heads flips before the second tails flip.

$X\sim\text{NegativeBinomial}(r=2,p=\frac{1}{2})$

We want
$ \begin{align*} P(X\geq4)=1-P(X\leq3)&=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)\\ &=1-\binom{0+1}{1}\frac{1}{2}^{2}\frac{1}{2}^0-\binom{1+1}{1}\frac{1}{2}^{2}\frac{1}{2}^1-\binom{2+1}{1}\frac{1}{2}^{2}\frac{1}{2}^2-\binom{3+1}{1}\frac{1}{2}^{2}\frac{1}{2}^3\\ &=1-\frac{1}{2}^{2}-2\cdot\frac{1}{2}^{3}-3\cdot\frac{1}{2}^{4}-4\cdot\frac{1}{2}^{5}\\ &=\frac{3}{16} \end{align*} $

Hopefully the makes sense. There are several equivalent ways of defining a negative binomial and writing the PMF so if the PMF is unfamiliar the way I wrote it I’m using:
$P(X=x)=\binom{x+r-1}{r-1}\cdot p^r\cdot (1-p)^{x}$

Academic Actuary · #4 12-11-2017, 02:01 AM

Quote:

Originally Posted by Z3ta

Let $X$ be the number of heads flips before the second tails flip.

$X\sim\text{NegativeBinomial}(r=2,p=\frac{1}{2})$

We want
$ \begin{align*} P(X\geq4)=1-P(X\leq3)&=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)\\ &=1-\binom{0+1}{1}\frac{1}{2}^{2}\frac{1}{2}^0-\binom{1+1}{1}\frac{1}{2}^{2}\frac{1}{2}^1-\binom{2+1}{1}\frac{1}{2}^{2}\frac{1}{2}^2-\binom{3+1}{1}\frac{1}{2}^{2}\frac{1}{2}^3\\ &=1-\frac{1}{2}^{2}-2\cdot\frac{1}{2}^{3}-3\cdot\frac{1}{2}^{4}-4\cdot\frac{1}{2}^{5}\\ &=\frac{3}{16} \end{align*} $

Hopefully the makes sense. There are several equivalent ways of defining a negative binomial and writing the PMF so if the PMF is unfamiliar the way I wrote it I’m using:
$P(X=x)=\binom{x+r-1}{r-1}\cdot p^r\cdot (1-p)^{x}$

Using this formula, you would have to sum x from 4 to infinity.

Z3ta · #5 12-11-2017, 02:06 AM

Quote:

Originally Posted by Academic Actuary

Using this formula, you would have to sum x from 4 to infinity.

Right or take the probability of the complement and subtract it from 1 as my calculation shows.

My guess is that the writer of the problem intended for it to be solved your way, but this is the general solution if it weren’t so easy to just list out the possibilities.

ActuaryKen5 · #6 12-11-2017, 11:19 PM

Quote:

Originally Posted by Academic Actuary

Four heads on the first four flips, or three heads on the first four with a head on the fifth.

Why do we have to separate it out like that?
Why wouldn't cant we just have HHHHT permuted 5 ways?

ActuaryKen5 · #7 12-11-2017, 11:22 PM

Quote:

Originally Posted by Academic Actuary

Using this formula, you would have to sum x from 4 to infinity.

If I use negative binomial I will still get 5/32.

Gandalf · #8 12-11-2017, 11:33 PM

Quote:

Originally Posted by ActuaryKen5

Why do we have to separate it out like that?
Why wouldn't cant we just have HHHHT permuted 5 ways?

That would miss HHHHH.

Z3ta · #9 12-11-2017, 11:36 PM

Quote:

Originally Posted by ActuaryKen5

If I use negative binomial I will still get 5/32.

Perhaps show your work?

Vorian Atreides · #10 12-11-2017, 11:43 PM

Quote:

Originally Posted by ActuaryKen5

Why do we have to separate it out like that?
Why wouldn't cant we just have HHHHT permuted 5 ways?

The bolded is irrelevant if you get 4 H first (because HHHHH would also satisfy the conditions, which you've not accounted for if considering 5 flips).

The probably of the first 4 being H is $0.5^4$

Now, you look at permuting HHHT for what takes place in the first 4 flips:
${}_4C_1\times 0.5^3 \times 0.5$

However, there is only a 0.5 chance that the fifth flip is H, so the above is multiplied by 0.5

Since the above two are mutually exclusive, we get the total probability as being their sum:

$0.5^4 + {}_4C_1\times0.5^3\times0.5\times0.5 = \frac{1}{2^4} + \frac{4}{2^5} = \frac{6}{2^5} = \frac{3}{16}$

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