probability

Flip it 1000 times and see what the results are. If you get 945 heads and 55 tails, you know its biased. The probability of getting a heads would be how many heads you got out of the total number of flips... Other than that, you'd probably need to know quite a bit of physics and a lot about the coin in question to figure this out..
 
Thbaks for the response.

because here is the problem i want to solve...


In a biased coin,the proability of head is 0.6 . Find the probability of 2 heads in 3 tosses .

a)0.288
b)0.432
c)0.375
d)0.144
 
\(\displaystyle \L
\frac{{3!}}{{1!2!}}\left( {.6} \right)^2 \left( {.4} \right)^1\)
 
My solution:

In toss-1 probaility of head P1 =0.6

In toss-2 probaility of head P1 =0.6

so 2 head achieved.

In toss-3 no head...but tail is expected .....tail probability P3=1-0.6=0.4


I am confused now.....
 
pka said:
\(\displaystyle \L
\frac{{3!}}{{1!2!}}\left( {.6} \right)^2 \left( {.4} \right)^1\)
Click to expand...

Hi,

what is this formula ?

is it appicable ONLY for biased coin ?

please let me know the actual formula so that i can understand it.
 
tkhunny said:
It's a whole discussion, possibly of some length. Look up the "Binomial Theorem".

http://mathworld.wolfram.com/BinomialTheorem.html
Click to expand...

Sir,

i know by nomial theorem...its used when u are going to expand any power. e.g (x+100)^12

but how binomial theorem is coing in this picture ?

i am completely confused .....i cant match all these things and cant not come to a conclusion.


Do u mean there is some formula for biased coin which in turn takes help of binomial theorem to find out the value ?

well. whats that formula then ?

thanks
 
n = # of flips
p = Probability of Heads on any one toss
1-p = q = Probability of Tails on any one toss

Expand: (p+q)^n

n = 1 ==> p + q
n = 2 ==> p^2 + 2*p*q + q^2
n = 3 ==> p^3 + 3*(p^2)*q + 3*p*(q^2) + q^3
 
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