The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find `P(X ge 1)`.

If the mean and variance of a binomial distribution are 4 and 4//3 . Then the distribution is

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

If the mean and variance of a binomial distribution are 4 and 2 respectively , then the probability of 2 successes of that binomial variate X , is

If the mean and variance of a binomial variate X are 8 and 4 respectively then P(X lt 3)=

The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. Then P(x gt 6)=

If the mean and variance of a binomial variable X are 2 and 1 respectively, then P(X ge1)=

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X=1) is Theoretical Distributions

If the mean and variance of a binomial distribution are 15//4 and 15//16 then the number of trials is

The mean and variance of a Binomial variate are 2.4 and 1.44 respectively. Find the parameters, P(X=2) and P(1 lt X le4) .