Question

After 5 coinflips, what is the probability that there are 3 heads?

Answer 1

I think it is \[\frac{5!}{3!(5-3)!}\times(\frac{1}{2})^5\]but why?

Answer 2

so there are \[\left(\begin{matrix}5 \\ 3\end{matrix}\right)\]combinations where I get 3 heads, and for each of those combinations the probability is 0.5^5 I get it now :-D

Answer 3

Write down outcomes of five coin tosses like this:
five tails: ttttt
2 heads, 2 tails then a head: hhtth
etc.

Then the probability of 3 heads is the probability of any of the possible outcomes with 3 heads:
that is, hhhtt, hhtht, hhtth, htthh, ...

Now how many outcomes are there like this? There are 5 choose 3, 5C3 = 5!/(3! (5-3)!) = 5!/(3! 2!)

Answer 4

and each of those has probability (1/2)^5

Answer 5

yay thanks

Answer 6

you're like the next Srinivasa Ramanujan

Answer 7

The othe rway to think about this let H = probability of a head, T = probability of a tail.

Then
(H + T)^5
gives all the possibilities of # heads and tails and their probabilities, and we want the term H^3.T^2.

By the binominal theorem, it has coefficient 5C3 hence the probability of 3 heads is

\[{5 \choose 3} H^3 T^2 = {5 \choose 3} (1/2)^5\]

Answer 8

This might seem slightly over the top notation for just heads and tails, but it's very helpful for other problems. For example: suppose we throw a fair six-sided die seven times. What's the probability we get 5 or 6 three times.

Well the outcome we want, 5 or 6, has probability P = 1/3, and the other outcome has probability Q = 2/3
Hence the probability of three 5 or 6 is\[{7 \choose 3} P^3Q^4 = {7 \choose 3} (1/3)^3 (2/3)^4\]

Answer 9

Great stuff. I feel enlightened :-D