If X follows Binomial distribution with mean 3 and variance 2 , then P(X≥8) is equal to :1739183919392039

The question seems to be a bit confused. If X follows a binomial distribution with a mean of 3 and a variance of 2, we can find the parameters of the binomial distribution. The mean of a binomial distribution is np and the variance is np(1-p), where n is the number of trials and p is the probability of success.

From the mean, we have np = 3. From the variance, we have np(1-p) = 2.

We can solve these two equations to find the values of n and p.

However, the number you provided, 1739183919392039, doesn't seem to fit into this context. It's unclear what this number is supposed to represent.

As for the probability P(X≥8), once we have the values of n and p, we can use the cumulative distribution function (CDF) of the binomial distribution to find this probability. However, without the correct values of n and p, it's impossible to find this probability.

Please provide more context or check the numbers in your question.