The mean and variance of a Binomial distribution are 4 and 4/3, respectively. Find the probability of more than two successes.a.433/729b.525/729c.656/729d.713/729

The mean and variance of a binomial distribution are related to the parameters n (number of trials) and p (probability of success on each trial) as follows:

Mean = np Variance = np(1-p)

Given that the mean is 4 and the variance is 4/3, we can set up the following equations:

4 = np 4/3 = np(1-p)

We can solve these equations simultaneously to find the values of n and p. From the first equation, we get p = 4/n. Substituting this into the second equation, we get:

4/3 = n*(4/n)*(1 - 4/n) 4/3 = 4 - 16/n n = 12

Substituting n = 12 into the first equation, we get p = 4/12 = 1/3.

The probability of more than two successes is 1 minus the probability of 0, 1, or 2 successes. We can calculate these probabilities using the binomial probability formula:

P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where C(n, k) is the number of combinations of n items taken k at a time.

P(X = 0) = C(12, 0) * ((1/3)^0) * ((2/3)^12) = (2/3)^12 P(X = 1) = C(12, 1) * ((1/3)^1) * ((2/3)^11) = 12 * (1/3) * (2/3)^11 P(X = 2) = C(12, 2) * ((1/3)^2) * ((2/3)^10) = 66 * (1/9) * (2/3)^10

Adding these up and subtracting from 1 gives the probability of more than two successes:

1 - [(2/3)^12 + 12*(1/3)(2/3)^11 + 66(1/9)*(2/3)^10] = 656/729

So, the correct answer is c. 656/729.