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#1
12-10-2017, 09:50 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5
Need help!

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?

Can someone help me out on this?
Much appreciated!
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#2
12-10-2017, 10:24 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,813

Four heads on the first four flips, or three heads on the first four with a head on the fifth.
#3
12-11-2017, 12:28 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

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:
$P(X=x)=\binom{x+r-1}{r-1}\cdot p^r\cdot (1-p)^{x}$

Last edited by Z3ta; 12-11-2017 at 12:39 AM..
#4
12-11-2017, 01:01 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,813

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.
#5
12-11-2017, 01:06 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

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.
#6
12-11-2017, 10:19 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5

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?
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#7
12-11-2017, 10:22 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5

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.
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#8
12-11-2017, 10:33 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,881

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.
#9
12-11-2017, 10:36 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by ActuaryKen5 If I use negative binomial I will still get 5/32.
#10
12-11-2017, 10:43 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: Hitler's Secret Bunker Studying for ACAS College: Hard Knocks Favorite beer: Sam Adams Cherry Wheat Posts: 62,089

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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