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i've been studying the following example for a couple of hours now, can someone explain/walk me through it

whats giving me the most trouble is (5 choose 2,2,1) and (5 choose 3,1,1) can someone explain those two binomial coefficients withing the context of the problem


A warped coin has probability of 0.5 of landing Heads, probability of 0.4 of landing Tails, and probability 0.1 of landing on its Edge. (I said it was warped!). It is flipped 5 times. What is the probability that more Heads occur than Tails?

Solution: We can split this problem into sub cases according to the number of Edges that occur:

No Edges (5, 4 or 3 Heads): (.5)5 +(5 choose 1)(.5)4(.4)+(5 choose 2)(.5)3(.4)2

One Edge (4 or 3 Heads): (5 choose 1)(.5)4(.1) + (5 choose 3,1,1)(.5)3(.4)(.1)

Two Edges (3 or 2 Heads): (5 choose 2)(.5)3(.1)2 + (5 choose 2,2,1)(.5)2(.4)(.1)2

Three Edges (2 Heads): (5 choose 2)(.5)2(.1)3

Four Edges (one Head): (5 choose 1)(.5)(.1)4

For the final answer we sum up all these values

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$5\choose 3,1,1$ is the number of ways you can have $3$ heads, $1$ edge, and $1$ tail. –  Gerry Myerson Feb 25 '13 at 5:28
    
got it! thanks. –  notamathwiz Feb 25 '13 at 5:35
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1 Answer 1

It may be easier to split another way. The probability of heads is $0.5$. So the probability of $3$ heads or $4$ or $5$ is $\binom{5}{3}(0.5)^5+\binom{5}{4}(0.5)^5+\binom{5}{5}(0.5)^5$.

We need in addition the probability of $2$ heads, and $0$ or $1$ tails, and therefore $3$ or $2$ edges, which is

$\binom{5}{2,0,3}(0.5)^2(0.4)^0(0.1)^3+\binom{5}{2,1,2}(0.5)^2(0.4)^1(0.1)^2$.

We need also the probability of $1$ head, $0$ tails, and $4$ edges, which is $\binom{5}{1,0,4}(0.5)^1(0.4)^0(0.1)^4$.

Remark: In general, suppose that we toss the coin $n$ times. We will find the number of toss patterns that result in $a$ heads, $b$ tails, and $c$ edges, where $a+b+c=n$.

The locations of the heads can be chosen in $a$ ways. For each choice of locations of heads, the locations of the tails can be chosen in $\binom{n-a}{b}$ ways. So the total number of patterns is $\binom{n}{a}\binom{n-a}{b}$. This is equal to $\frac{n!}{a!(n-a)!}\cdot \frac{(n-a)!}{b!(n-a-b)!}$.

Cancel the $(n-a)!$ from numerator and denominator, and note that $n-a-b=c$. Thus the number of patterns is $\frac{n!}{a!b!c!}$. This is precisely $\binom{n}{a,b,c}$.

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