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probability Problem

 
 
Reply Tue 29 Mar, 2011 06:51 pm
The problem is
Find the probability when have at least 2 heads in 3 tosses of a coin
I can solve it by the list. But is's too slow if I have a big number

Like, The probability when have at least 3 heads in 5 tosses
Somebody solve it, by use combination like 5C3/(1/2)^5

But I'm so confused, the we use conbination if the position of heads and tails are not important. But here, the position is important.

Why everybody solve it by using the combination
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Type: Question • Score: 1 • Views: 967 • Replies: 3
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Oylok
 
  1  
Reply Tue 29 Mar, 2011 08:47 pm
@daivinhtran,
daivinhtran wrote:
Somebody solve it, by use combination like 5C3/(1/2)^5


That should be 5C3 / 2^5.

Quote:
here the position is important.


Yes, it is. Do you know what that "5C3" means?

Let's use H for "Heads." Let's use T for "Tails". Every result of five tosses can be written to look like something this: HTTHH. (Here you first get heads, then 2 tails, then 2 more heads.)

000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1 2 33 4 5
Another way to think about it, is that we are choosing different places for the H's. If we choose places 1, 4 & 5, then we get HTTHH. If we choose places 2, 3 & 4, then we get THHHT. The heads are in different positions, which is important.

Every choice of three numbers (like the combination 1, 4 & 5) gives you a different result (like HTTHH). You can choose 3 numbers from 5 numbers in 5C3 different ways. Therefore, you have 5C3 different possible ways to get 3 H's in 5 tosses.

Quote:
I'm so confused, the we use conbination if the position of heads and tails are not important.


No, that's not correct.

The order of the heads and tails matters.

Here is what doesn't matter: it doesn't matter what order you put the three numbers in. So if you put the heads in places 1, 4 & 5; that is the same as putting them in places 1, 5 & 4. Either way you get HTTHH.
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Oylok
 
  1  
Reply Tue 29 Mar, 2011 08:51 pm
@daivinhtran,
daivinhtran wrote:

Like, The probability when have at least 3 heads in 5 tosses
Somebody solve it, by use combination like 5C3 / 2^5

(The red words above are mine.)

Also, your formula is the right one for exactly 3 heads.

It is not correct if you're looking for at least 3 heads.
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oolongteasup
 
  1  
Reply Tue 29 Mar, 2011 10:08 pm
@daivinhtran,
Whenever I have two heads or more I like to combine them!

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

There are 2 possibilities in each of 5 tosses so there are 2^5 possible outcomes

5HEADS IS 5C5 / 2^5 = 1/32

4HEADS IS 5C4 / 2^5 = 5/32

3HEADS IS 5C3 / 2^5 = 10/32

2HEADS IS 5C2 / 2^5 = 10/32

1 HEAD IS 5C1 / 2^5 = 5/32

0 HEAD IS 5C0/ 2^5 = 1/32

For further reference wiki Permutation


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