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 formula for coin toss
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 10/16/2012 :  11:50:12  Show Profile  Reply with Quote
It has been a long time since I have used C(n,r). I would like to know how to apply it to the following two problems:
1. Flip a coin 3 times, what is the probability for 2 heads to come up (order doesn't matter)?
2. Flip a coin 3 times, what is the probability for the 1st and 2nd toss to be heads and then the 3rd be tails?
By listing out HHH HHT HTH HTT THH THT TTH TTT
I get the answer to #1 of 3/8 and #2 to be 1/8
However, what is the "formula" to get the answers?
I am thinking #1 is C(3,2)/8 but that doesn't really look like a formula because of the "8". If I were to use a generic formula, what would it be. Something like C(n,r)/(2^n) ??
I don't know of a 'formula' that works for #2.
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 10/16/2012 :  21:29:39  Show Profile  Reply with Quote
Since there are two choices for each of the three tosses, there are 2 or 8 possible results. Now if you want 2 heads, the third must be a tail. There are 3 ways to get one tail, i.e. THH, HTH, or HHT. (notice the pattern). This can also be shown as C(3,2) or 3 ways. Now for the last one, there is only one way for a HHT to occur because all three are fixed. This can also be shown as C(3,3) or
C(3,0) or one way. So to find the probabilies it's just C(3,2)/2 or C(3,3)/2
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 10/18/2012 :  16:01:14  Show Profile  Reply with Quote
Thanks.
I would like an explanation of what C(n,r) means in this context (coin toss).
C(n,r) means there are "n" different items and C(n,r) gives the number of combinations of these n items "r" at a time.
However, what does C(3,2) for THH HTH HHT mean?
There are not "3" different items and even so, they aren't being taken "2" at a time.

So, to figure out getting 5 heads in 8 tosses would be C(8,5)/(2^8)?
If so, I understand that but still don't see how the "n" and "r" are described.
Further, let's say there is a three sided coin. "H"eads, "T"ails and "O"ther. How would you figure getting 5 heads in 8 tosses?
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 10/30/2012 :  09:44:56  Show Profile  Reply with Quote
Let's say there is a three sided coin. "H"eads, "T"ails and "O"ther. How would you figure getting 5 heads in 8 tosses?
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 11/14/2012 :  10:56:18  Show Profile  Reply with Quote
Okay, I suppose no one is going to answer the following:
quote:
Originally posted by the_hill1962

Let's say there is a three sided coin. "H"eads, "T"ails and "O"ther. How would you figure getting 5 heads in 8 tosses?


After thinking about it, I guess it was a stupid question. The C(n,r) only is for "binomial" types of problems (am I correct in saying "binomial" for the reason that flipping a coin is h or t?).
However, is there not a way to make a formula for
Let's say there is a three sided coin. "H"eads, "T"ails and "O"ther. How would you figure getting 5 heads in 8 tosses?
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 11/15/2012 :  12:08:43  Show Profile  Reply with Quote
Assuming each outcome is equally likely, there are 3^8 or 6561 outcomes. The only way I can see how to do it (although I think there must be a better solution) is using a tree diagram, but that's too much work. I do the same thing when we have our hockey playoffs but I wait until there are only 3 games or 27 possibilities left. (Ties are not broken.) That's manageable.
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royhaas
Moderator

USA
3059 Posts

Posted - 11/16/2012 :  08:10:14  Show Profile  Reply with Quote
The generalization of C(n,r)is n!/(r1!...rk!) where r1+...+rk=n. In other words you can have k>2 outcomes instead of just 2. One way to attack the specific problem is to consider the expansion of (H+T+O)^8. Add the coefficients of all terms containing H^5. Note that the sum of the exponents of T and O in this case must be 3. Recall that C(n,r) comes from considering (H+T)^n. The multinomial generalization is straightforward, although a little more complicated than the binomial case.
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 11/16/2012 :  18:00:21  Show Profile  Reply with Quote
Oh my, what you forget in 5 years, lol. Yes, you write it as (H+[T+0])^8 and find the coefficient of H^5.
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 11/17/2012 :  15:23:57  Show Profile  Reply with Quote
Continuing the thread, I have a bit more time to write.
Remembering Pascal's triangle for the expansion of (a+b)^n, the first four are:
a+b
a+2ab+b
a+3ab+3ab+b
a^4+4ab+6ab+4ab+b^4
Now consider the coefficients of the last line 1,4,6,4,1 which add up to 16. This idea is frequently used in bridge to determine the possiblities of a split. So imagine that there are 4 trump out and you want to look at the splits. The a represents a trump card on the left and b the right. So the number of cases where there is a 4-0 split is 1, the first coefficient. The number of cases where there is a 3-1 split, is 3, a 2-2 split 6, a 1-3 split 4 and a 0-4 split 1. There are then 16 cases, so the chance on average of a 2-2 split is 6/16, in other words, the odds are always against it. This is obviously an application of the binomial theorem. So to find the number of heads when you have H,T, or O, for 8 trials is as was suggested above, you apply the binomial theorem for (H+[T+O])^8, with the H behaving as an 'a' and the [T+O] behaving as a 'b'. Once that is done, you will have to expand the [T+O], [T+O], [T+O], and so on...
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 11/26/2012 :  12:37:19  Show Profile  Reply with Quote
Now I feel better. It has been 15 years since I have had to work this type of problem.
Thanks Ultraglide and royhass!
quote:
Originally posted by Ultraglide

Oh my, what you forget in 5 years, lol. Yes, you write it as (H+[T+0])^8 and find the coefficient of H^5.

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