Probability

[dcc3038026]

What is the probability that 14 tosses of a fair coin will show 6 heads?

 

[arthur ohlsten]

let h represent a head=1/2
let t represent a tail=1/2

[h+t]^1=h+t this is one toss
[h+t]^2=h^2+2ht +t^2 this is two tosses
h^2 is 2 heads 1/4 , 2ht is one head one tail 2/4
t^2 is two tails 1/4

[h+t]^14 = h^14 + 14 h^13 t^1 + [14x13]/2h^12 t^2 +...14C6 h^6t^8+... t^14
but 6 heads = 14!/[6!x8!] [1/2]^14

i have no computer so determining a value I will leave to you.
Arthur

 

[soroban]

Hello, dcc3038026!

What is the probability that 14 tosses of a fair coin will show 6 heads?

\(\displaystyle \text{Answer: }\;{14\choose6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^8\)

 

[arthur ohlsten]

I assume the student is familiar with the notation used for the combination of 14 things taken 6 at a time [14,6]

To play safe I wrote it as 14! /[6! 8!]
I also tried to show a method of expanding a math term for 14 tosses,i.e. [P+Q]^14

Arthur

 

[Denis]

but 6 heads = 14!/[6!x8!] [1/2]^14

Arthur, shouldn't that be 2^14, not (1/2)^14 ?

 

[arthur ohlsten]

no
leaving off the coefficients:
the probability of 1 head is 1/2
the probability of two heads is [1/2]^2=1/4

the probability of 6 heads and 8 tails=[1/2]^6[1/2]^8
or 6 heads and 8 tails is [1/2]^14

the probability of 6 heads and 8 tails for 14 tosses
14C6 [1/2]^6 [1/2]^8 =3003[1/2]^14
3003/16384=.1833 answer

please check the arithmetic for errors

Arthur

 

[Deleted member 4993]

The problem does not say "exactly 6 heads".

When we have 7 heads (or 8, or 9...), we have 6 heads showing also

 

[Denis]

> but 6 heads = 14!/[6!x8!] [1/2]^14

Ok Arthur; I thought the square brackets meant the whole thing was the divisor...

 

[mmm4444bot]

The problem does not say "exactly 6 heads" ...

Agreed! There is plenty of ambiguity in the world already, in general; the disicpline of statistics, in particular, is one of the worst places to add more.

Either it's exactly six or at least six.

Cheers,

~ Mark

 

[arthur ohlsten]

If it is the probability of 6 or more for 14 tosses the answer is:
where P[n] stands for the probability of n heads
P[6]+P[7]+P[8]+P[9]+P10]+P[11]+P[12]+P[13]+[P14]
OR
1-{P[0]+P[1]+P[2]+P[3]+P[4]+P[5]}

where P[n]=nC14 [1/2]^14
nC14 is defined as the combination of 14 things taken n at a time
14!/{n![14-n]!}

Or you can obtain the coefficients by means of Pascals Triangle.
or you can obtain the coefficients by expanding [P+Q]^14

Arthur