A coin toss probability problem

[#neutrino]

Homework Statement

If a fair coin is tossed 6 times what is the probability of obtaining 3 heads and 3 tails /

Homework Equations

c(n,r)=n!
-------
r!(n-r)!

The Attempt at a Solution

the no of probable outcomes = 2^{6
= 64
here's what i can't get around how to calculate the no. of times to get 3 heads and 3 tails out of all the other outcomes .}

 

[anachin6000]

If you have to calculate something like this you could combine the probabilities. If you have to get the result X in the first experiment and the result Y in the second, the total probability is the product of the particular probabilities (only when the first experiment does not influence the second). also, the probability of getting the result the result X or the result Y in one experiment is the sum of the particular probabilities.(You should try to prove those theorems if you haven't herd of them before.)
In your problem, the probability of getting 3 heads and 3 tails is 1/2⁶, but this result can be obtained in various combinations. The number of those ways to combine the results is the number of ways in which you can choose 3 experiments out of 6 to get one particular result. That's C₆³.
Notice that the number you are asking for is C₆³ and using your method you get the same result.

 

[#neutrino]

i still don't really get why we have to choose 3 experiments out of 6

 

[anachin6000]

i still don't really get why we have to choose 3 experiments out of 6

Suppose you have 6 boxes numbered from 1 to 6. Each time you toss the coin you write the result on a paper and put it in the connected box (the first result in box 1, the second in box 2, etc.). You aim to have three papers on which is written tails and three on which is written heads.
Now if you get three papers of each result, you might choose for example tails. Put all the papers with tails in the boxes. This way you get a way of combining your 3-3 result, because the empty boxes will get the rest of the papers (the ones with heads). So it is enough to put one set of papers to get a solution. that's why you can choose just 3 boxes(experiments) out of 6.
You can think the same for results like 2 tails- 4 heads. This way you can observe that the number you get from choosing 2 experiments out of 6 or 4 out of 6 gives the same result. So you will practically prove that C₆² = C₆⁴.

 

[#neutrino]

yup thnx got it

 

[Ray Vickson]

Suppose you have 6 boxes numbered from 1 to 6. Each time you toss the coin you write the result on a paper and put it in the connected box (the first result in box 1, the second in box 2, etc.). You aim to have three papers on which is written tails and three on which is written heads.
Now if you get three papers of each result, you might choose for example tails. Put all the papers with tails in the boxes. This way you get a way of combining your 3-3 result, because the empty boxes will get the rest of the papers (the ones with heads). So it is enough to put one set of papers to get a solution. that's why you can choose just 3 boxes(experiments) out of 6.
You can think the same for results like 2 tails- 4 heads. This way you can observe that the number you get from choosing 2 experiments out of 6 or 4 out of 6 gives the same result. So you will practically prove that C₆² = C₆⁴.

The usual notation (at least in North America) is $C_4^6$, not $C_6^4$, although ${}_6C_4$ or ${6 \choose 4}$ are preferred.

 

[anachin6000]

The usual notation (at least in North America) is $C_4^6$, not $C_6^4$, although ${}_6C_4$ or ${6 \choose 4}$ are preferred.

Well, I'm from Europe. But is nice to learn worldwide notations:)