Question

Anyone good with probability using combinations ?

Answer 1

@chmE

Answer 2

a coin is tossed ten times in a row, what is the probability that heads comes up exactly 5 times ?

Answer 3

so it doesn't matter what order heads comes up in as long as it is 5 times. This is conditional that tails is also 5 times... I need to think, I had stats 3 years ago

Answer 4

yes doesn't matter, and oh cool, this is precalculus lol

Answer 5

take ur time, appreciate it, thanks so much.

Answer 6

yeah this is just a small topic we're learning, which is combinations, we learn a little bit of everything in precalc lol.

Answer 7

well this is what I did but got the wrong answer, c(10,5)*c(10,5) / c(20,10) it should be .246 but I got .343 =/

Answer 8

\[\left(\begin{matrix}10 \\ 2\end{matrix}\right)=\frac{ 10! }{ 8!2! }=45\]=total possible arrangements of heads and tails. Now I have to think about what goes on top

Answer 9

I can't think of a cool math way, but we can draw it out

hhhhhttttt
hhhhttttth
hhhttttthh
hhttttthhh
httttthhhh
ttttthhhhh
tttthhhhht
.... and so on til 10

10/45= 2/9 is my answer

Answer 10

hmm... it was supposed to be done doing combinations, do you know how to do those ?

Answer 11

wait im sorry.... it's not combinations, actually the bionomial theorem, MY BAD!!

Answer 12

http://en.wikipedia.org/wiki/Binomial_theorem

There is combinations in this theorem, scroll half way down.

I'm going to have to jump back to my own problem now, hopefully somebody can give a nice formula

Answer 13

a coin is tossed ten times in a row, what is the probability that heads comes up exactly 5 times ?

First note we can count the total number of possible coin flip results rather easily; note we have \(2\) possibilities for each flip, and \(10\) flips in all, so by the rule of product aka fundamental counting principle we know we have \(\underbrace{2\times2\times2\times\dots\times2}_{10\text{ times}}=2^{10}=1024\) possible results. Cool!

Next, we want to count how many ways we could get exactly \(5\) heads out of \(10\) coin flips; this is just the number of ways to pick a combination \(5\) objects out of a pool of \(10\), i.e. \(_{10}C_5=\dbinom{10}5=\dfrac{10!}{(10-5)!}=\dfrac{10!}{5!}=10\times9\times8\times7\times6=252\)
So our probability is then just the ratio \(252/1024\approx0.246\)

Answer 14

Alternatively, recognize the probability of getting \(5\) heads and thus \(5\) tails, assuming a fair coin where both outcomes are equally likely per coin flip and every coin flip's result is independent of the others, is just the product of the probabilities per coin flip result i.e. \(\underbrace{\left(\dfrac12\right)\left(\dfrac12\right)\cdots\left(\dfrac12\right)}_{10\text{ times}}=\left(\dfrac12\right)^{10}=\dfrac1{2^{10}}=\dfrac1{1024}\). So we know the probability of any such result of exactly \(5\) heads is \(\dfrac1{1024}\), now we just have to count the number of ways of getting such a result -- which is just the number of combinations of \(5\) coin flips out of our total of \(10\): $$\dbinom{10}5=\frac{10!}{(10-5)!}=\frac{10!}{5!}=10\times9\times8\times7\times6=252$$For the probability that *any* of these mutually exclusive events could occur, we merely add them all:$$\underbrace{\frac1{1024}+\frac1{1024}+\dots+\frac1{1024}}_{252\text{ times}}=\frac{252}{1024}\approx0.246$$

Answer 15

Oh wow, thanks so much, appreciate it, most helpful person I have ever met on here lol