. A random variable has a mean value of 5 and variance of 3. What is the least value of Prob [|X - 5 < 3).

The question seems to be asking for the probability that the random variable X deviates from its mean (5) by less than 3. In mathematical terms, this is P(|X - 5| < 3), which is the same as P(2 < X < 8).

Without knowing the specific distribution of the random variable X, we can't calculate this probability exactly. However, we can use Chebyshev's inequality to get a lower bound on this probability.

Chebyshev's inequality states that the probability that a random variable deviates from its mean by k standard deviations is at most 1/k^2. In this case, the standard deviation is the square root of the variance, which is sqrt(3) ≈ 1.73. The value of 3 is approximately 1.73 standard deviations away from the mean.

So, using Chebyshev's inequality, we have P(|X - 5| < 3) ≥ 1 - 1/(1.73^2) ≈ 0.334.

Therefore, the probability that the random variable X deviates from its mean by less than 3 is at least 0.334.