determine prob. of getting 8 heads after tossing 10 coins

[pc678]

Say I had 10 coins.

How do I determine the probability of getting 8 heads?

 

[arthur ohlsten]

Re: Tossing coins

[p+q]^10=1 is all possible outcomes of ten tosses where p=head and q=tails
p^10q+10p^9q^1 + [10[9]/2]p^8q^2 + ... all the way down to q^10=1

thevcoefficients of the terms are the Pascal triangle terms

probability of 8 heads is 8C10 p^8q^2
probability of at least 8 heads 8C10 p^8q^2 + 9C10 p^9 q^1 + 10C10 p^10

where 8C10= 10!/[8![10-8]!]
Arthur

 

[pc678]

Re: Tossing coins

I'm at a bit of a loss. Can you tell me more about the equation you used? Does the equation you used come out to 45? I don't really know how to convert it to a p-value.

Would I also use 10! to determine how many ways I can arrange how 8 heads could occur? I know that would be a ridiculously large number.

 

[arthur ohlsten]

Re: Tossing coins

let us take a easier problem
what is the probability of 2 heads if we toss a coin twice?

let P = probability of a head [1/2]
let q be the probability of a no head, a tail[1/2]

[p+q]^2=1 is all possible outcomes. Expanding
1=p^2+2pq+q^2

p^2=probability of 2 heads
pq= probability of 1 head 1 tail
q^2=probability of 2 tails

prob of 2 heads=p^2
p^2=1/4

prob of 1 head 1 tail=2pq
2pq=2[1/4]=1/2

prob of 2 tails=q^2
q^2=1/4

the coefficients 1,2,1 can be determined by pascals triangle [ if you are familiar with it]
the coefficients can be obtained from the cobination terms[ nCr= n!/[n!(n-r)!]

2C1= 2!/[1!(2-1)!]
2C1=2

2C2= 2!/[2!0!]
2C2=1

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Now for the combination of tossing a coin 10 times
1=[p+q]^10
this will have all combinations
1=p^10+10p^9q+{10[9]/2}p^8q^2 .... q^10

as P exponent decreases q exponent increases so sum of exponents always equal 10
the coefficient of the terms are from pascals triangle or from the combinations

I shall do the coefficient for the combination of 10 things taken 8 at a time,[ our coefficient of 8 heads and 2 tails].
10C8=10!/[8!(10-8)!]
10C8=10!/[8!2!]
10C8= 10(9)8!/[8!2!]
10C8=10(9)/2!
10C8=10(9) / 1(2)
10C8=45
so probability of 8 heads= 45 [1/2]^8[1/2]^2
prob of 8 heads= 45[1/2]^10

to show the probability of 9 heads=10C9p^9q^1
prob={10/1} [1/2]^10
prob=10/1024

how did you know coefficient was 10/1 ?

let us look at 10!/{7!(10-7)!} the coefficint of 7heads
10!/(7!3!)
10x9x8 /[1x2x3]

the coefficients come out to have equal number of terms on top and bottom

hope all this helps

Arthur