If a coin is flipped 10 times, what is the probability that it will land heads-up at least 8 times? 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.
 A: 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 !
A: 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?
A: 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
