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I absolutely remember learning this is middle school, yet I cannot remember how to solve it for the life of me. Something to do with nCr, maybe? ...

Thanks for any help.

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You might want to look at the Binomial Distribution – Louis Jan 30 '12 at 10:40
P(at least 8 heads) = P(8 heads) + P(9 heads) + P(10 heads) = (10C8 + 10C9 + 10C10) / 2^10 – Chris Taylor Jan 30 '12 at 10:40
Assuming this is NOT homework, $P(\text{At least }8)= P(8) + P(9) + P(10)$ $P(x) \times 2^n= C^n_x = \frac{n!}{x!(n-x)!}$ For A Binomial Distribution $$(P(8) + P(9) + P(10))\times2^{10} = \frac{10!}{8!.2!} + \frac{10!}{9!.1!} + \frac{10!}{10!.0!}$$ $$=\frac{10.9}{2} + 10 + 1 = 56$$ – Inquest Feb 29 '12 at 14:00

4 Answers 4

What we'd like to do is find a way to set the problem up in some way that we know how to solve it.

$P($At least $8$ heads) = $P(X \geq 8)$ where $X$ is the Random Variable associated with the number of heads attained.

Well, since $X$ can only have the values $0$ through $10$, perhaps we should split $P$ up:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

We can split them up like this because there is no "overlap" between the events (You can't get 8 heads and then get either 9 or 10 heads too.)

Now we just need to apply the definition of probability:

$P(S) = n(E)/n(S)$

where $n(E)$ is the number of items in our event set, and $n(S)$ is the number of items in our sample space.

Well, for each of the probabilities, $n(S)$ = $2^{10}$ by the multiplication principle. Now, what are each of the $n(E)$?

You thought it would have to do with Combinations (nCr), and you were right. We use combinations instead of permutations because we really don't care which order we get the heads in, right?

So, for $X = 8$: $n(E) = $${10}\choose{8}$

and so on. Can you take it from here?

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One way you could get $8$ heads is $HHHHHHHHTT$ with probability $(1/2)^{10}$.

But the heads could occur in $C(10,8) = 10!/8!2!$ sequences, so $P[8H] = C(10,8)/2^{10}$

Similarly, $P[9H] = C(10,9)/2^{10}$ and $P[10H] = C(10,10)/2^{10}$

Add them up !

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Lets split it into cases: 8 heads, 9 heads, and 10 heads. there are C(10, 8) (10 choose 8)= 45 sequences with 8, C(10, 9)= 10 sequences with 9 heads, and of course 1 case with 10. 45+10+1= 56

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Considering there are two ways this can go, I view it as a True/False statement. Either you are going to flip heads, or you simply won't. You can figure it out the hard way, or use a formula. I start out with 2, since there are only two options. From then on, it branches out: Ex: 2^0 + 2^1 + 2^2 + 2^3 + etc. until 2^8 = 511.

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I don't understand what you are trying to say here. Also, $2^8 \neq 511$. – Michael Albanese Apr 8 at 2:36

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