- #1

JPC

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ok , the probability is obviously 50 : 50

but , what's the probability that your results are same as the probability for this case

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- Thread starter JPC
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- #1

JPC

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ok , the probability is obviously 50 : 50

but , what's the probability that your results are same as the probability for this case

- #2

radou

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The probability of *what* is '50 : 50' ?

- #3

JPC

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well

well if u throw 100 times a coin

the probability is : 50 heads, 50 tails

well if u throw 100 times a coin

the probability is : 50 heads, 50 tails

- #4

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Is this homework? If so, please show what you have done. If not, I (or someone else here) will show you how to compute such a probability.

- #5

ssd

- 268

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well

well if u throw 100 times a coin

the probability is : 50 heads, 50 tails

Not for any coin. Only for an unbiased coin P(Head)=P(Tail) in a single toss = 0.5, assuming that there is no other outcome of the toss.

Getting 50 heads and 50 tails is an event, it is not a probability.

Again, if P(H)=0.5 then

Now comming to your question: If you denote the no. of heads in 100 tosses by X, then X~Bin(100,0.5). Find P(X=50).

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- #6

JPC

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If you denote the no. of heads in 100 tosses by X, then X~Bin(100,0.5). Find P(X=50).

What does that mean , i don'k know all the conventions yet ?

and its not for homework

- #7

radou

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ssd was talking about random variables. You may want to google-up that term.

- #8

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[tex]\frac{100!}{50!\,50!}\;\frac 1 {2^{100}} \approx 0.0796[/tex]

- #9

ssd

- 268

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only ifthe coin is unbiased and tosses are independent. Assuming an unbiased coin and indepdendent tosses, the answer to the original question is

[tex]\frac{100!}{50!\,50!}\;\frac 1 {2^{100}} \approx 0.0796[/tex]

Right you are. I did not mention the fact that the sum of Bernoullian trials will be Binomial if those trials are independent. Actually I really was focused on the statement of the problem.

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- #10

JPC

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hey , u didnt put any operator between the 50! and 50!

is it multiply ?

and , u didnt put any operator between the two fractions, is it also multiply ?

because , if it is , doesn't look very much like 0.0796

100! : 5050

50! : 1275

50!² : 1625625

5050 / (1625625 * 2^100) = 2.45...

and, also, in general terms, what is the logic between this equation ?

is it multiply ?

and , u didnt put any operator between the two fractions, is it also multiply ?

because , if it is , doesn't look very much like 0.0796

100! : 5050

50! : 1275

50!² : 1625625

5050 / (1625625 * 2^100) = 2.45...

and, also, in general terms, what is the logic between this equation ?

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- #11

Parthalan

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I'm not that good at probability, but the vinculum affects the order of operations, i.e.

[tex]\frac{100!}{50!50!} \frac{1}{2^{100}} = ( 100! / 50!50! ) * ( 1 / 2^{100} ) = 0.0795892374[/tex]

[tex]\frac{100!}{50!50!} \frac{1}{2^{100}} = ( 100! / 50!50! ) * ( 1 / 2^{100} ) = 0.0795892374[/tex]

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- #12

CRGreathouse

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because , if it is , doesn't look very much like 0.0796

100! : 5050

50! : 1275

50!² : 1625625

5050 / (1625625 * 2^100) = 2.45...

Wha...? 50! is a huge number with 65 digits, it's a lot bigger than 1275. 1275 = 50 + 49 + ... + 2 + 1, while 50! = 50 * 49 * ... * 2 * 1.

- #13

JPC

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the ! is a multiplication , not addition, not 0.5(n² + n)

i see my mistake

- #14

JPC

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like what's the equation to find it

like if u want to find the sum of all the numbers from 0 to x its : 0.5(x + x²)

but what is it for the multiplication ?

- #15

CRGreathouse

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like what's the equation to find it

like if u want to find the sum of all the numbers from 0 to x its : 0.5(x + x²)

but what is it for the multiplication ?

There's no polynomial that gives a value close (for any reasonable definition of "close") to x! for all positive x.

There are some approximation formulas; (x/e)^x ~ x! would be one of the simplest*, where e = 2.71828...

If you want the exact answer you're just going to have to multiply it out, though. For this problem that's pretty much all you can do. Notice, though, that 100!/50! = 100 * 99 * ... * 51 * 50 * 49 * ... * 2 * 1 / (50 * 49 * ... * 2 * 1) = 100 * 99 * ... * 52 * 51.

* I hope I got this right, it's from memory. There are much better ones if you need more accuracy; Gosper's reformulation of Stirling's estimate is pretty good.

- #16

JPC

- 206

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better solution (if x is smaller than 1 million)

learn visual Basics , and make a program to do it for you using 'do until' - 'loop'

But what's the logic in that equation ?

how can i make an equation like that for other cases ?

whats the clear syntax ?

learn visual Basics , and make a program to do it for you using 'do until' - 'loop'

But what's the logic in that equation ?

how can i make an equation like that for other cases ?

whats the clear syntax ?

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