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Taking a Probability & Statistics class this term and trying to get my head wrapped around how I calculate coin tosses with specific out comes in mind. We're using the nCr and nPr functions on our calculators to speed the calculations along. $^nP_r = n(n-1)(n-2)\ldots(n-(r+1)$ and $^nC_r=\large \frac{n!}{r!(n-r)!} $

If I toss a coin 8 times and want to know the odds of heads coming up 3 times then I use $^8C_3$ for my total # of possible valid results over $2^8$. I use $^nC_r$ because its unordered.

However, if I do care that heads comes up 3 times at specific positions within the sequence or tails appears on the last toss, would I use $\large \frac{^8P_3 + ^8P_1}{2^8}$ or just $\large \frac{^8P_4}{2^8}$?


It appears that I misunderstood the method in general. The events were not exclusive of each other. The odds of heads coming up in the first three places OR tails coming up as the last is what was a win situation. That should give me $\large \frac{(2^5 + 2^1 - 2^3)}{2^8}$

I hope I got that correct. Anyone care to confirm for me?

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I don't know what a "win" is for your second question. I cannot think of any reasonable question about coin tossing in which we end up using the "permutation symbol". – André Nicolas Jul 21 '14 at 14:18
You ask: "If I do care that X or Y, ..." What do you want, X or Y or both? – user21820 Jul 21 '14 at 14:33
I think that total possibility will be 16 and not $2^8$ – MonK Jul 21 '14 at 15:33
up vote 1 down vote accepted

If I understand you well then you are dealing with two events. The first is that head turns up at $3$ specific positions. In your question it is unclear whether head is allowed to turn up at other positions or not. Let's say it is (if not then have a look at the answer of Eupraxis1981) and denote the event by $A$. The second is that tail appears on the last toss. Let's denote it by $B$. You are asking for $P\left(A\text{ or }B\right)$. Then formula: $$P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{ and }B\right)$$ can be practicized. This - if the coin is a fair one - with $P\left(A\right)=2^{-3}$ and $P\left(B\right)=2^{-1}$.

Note that it matters what the specific positions are. If the last toss is one of them then $P\left(A\text{ and }B\right)=0$ but in other cases $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)=2^{-4}$.

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