Question

You are given a bag of coins with varying biases.the probability of heads is a random variable sample from uniform distribution u[0,1].you draw a coin from these bag and toss it 100 times what is the probability of getting 100 heads stackover flow

Answer 1

Answer:

Step-by-step explanation:

Suppose I have a bag of n biased coins, and I sample without replacement m<n of them, and measure each coin i for their parameter pi∈[0,1], that is each coin is Bernoulli(pi). Now I would like to ask what is the most likely pm+1 on the next coin I pick. I am not sure how to answer this question aside from taking the mean of all m coins' parameter so far. That is: p^m+1=p1+…+pmm.

Answer 2

The probability of getting 100 heads when you toss the coin 100 times is ;

P(x = 100) = 7.8886 e^-231

let x = random variable = 100

x ~ U[0,1]   where U[0,1] = uniform distribution

number of tosses ( n ) = 100

mean of the uniform distribution = $\frac{0+1}{2} = \frac{1}{2}$

note;  the mean of a binomial distribution = np

∴ np = $\frac{1}{2}$

= 100 * p = $\frac{1}{2}$

hence ; p = $\frac{1}{200}$

next determine the probability of getting 100 heads ( i.e. P(x = 100 )

= nC x *P^x *(1−p)^n−x

= (  100C₁₀₀ ) * ( $\frac{1}{200}$ )¹⁰⁰ * ( 1 -

∴ P ( x = 100 ) = 7.8886 e^-231

Hence the probability of getting 100 heads stack over flow = 7.8886 e^-231

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