Calculating COV(A,B) for a Coin Throw Experiment

[proaction]

I have a problem with this question
can anyone help me please ..

We throw a coin with two sides, one has 1 on it and the other has 0 on it, the probability of getting 1 is 2/3 and 0 is 1/3 .

now we throw the coin N times , let A be the number of turns we got 0 on them , and be the number of turns we get 1 on them, what's the COV(A,B) ?

thanks

 

[chiro]

I have a problem with this question
can anyone help me please ..

We throw a coin with two sides, one has 1 on it and the other has 0 on it, the probability of getting 1 is 2/3 and 0 is 1/3 .

now we throw the coin N times , let A be the number of turns we got 0 on them , and be the number of turns we get 1 on them, what's the COV(A,B) ?

thanks

Hey there.

Looking at your problem you have a binomial distribution with p = 1/3 and N being the number of trials.

So basically X ~ Binomial(N,1/3)

Then we are given A = number of turns we get a 0 and B is number of turns we get a 1.

For N trials we have 2^N possibilities.

The distribution is going to range from having 0 matches to having N matches.

For distribution A we will use the parameter p = 2/3 which corresponds to the probability function being NCk (2/3)^(N-k) * (1/3)^k where k is the number of '0's we get.

B is simply a binomial distribution with p = 1/3 which corresponds to the probability function being NCk (1/3)^(N-k) * (2/3)^k.

So A ~ Binomial(N,2/3) and B ~ Binomial(N,1/3)

The definition of covariance is defined to be Cov(A,B) = E[AB] - E[A]E.

We know that E[A] = N*2/3 and E = N*1/3 so E[A]E = (N^2) * (2/9)

As for E[AB] we can use we formula for conditional expectation.