Question

​Three numbers are selected at random from first six positive integers. If X denotes smallest of the three numbers obtained, find the probability distribution of X. Also find mean and variance.

Answer 1

Total number of outcomes : N =  6C3 or C(6,3) = 6!/(3! 3!) = 20
set of outcomes:
     = { (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (1,3,6),
          (1,4,5), (1,4,6), (1,5,6), (2,3,4), (2,3,5), (2,3,6),
          (2,4,5), (2,4,6), (2,5,6), (3,4,5), (3,4,6), (3,5,6), (4,5,6) }

X = smallest number of the three numbers selected.

Probability distribution:
                  p(X= 0) = 0
N(X = 1) = 10        p(X = 1) = 10/20 = 1/2
N(X = 2) = 6          p(X = 2) = 6/20 = 3/10
N(X= 3) = 3           p(X = 3) = 3/20 
N (X = 4) = 1         p(X = 4) = 1/20
                         p(X >= 5)  = 0

Mean = μ = Σ X * p(X) = 1* 1/2 + 2 * 3/10 + 3 * 3/20 + 4 * 1/20
                = 35/20 = 1.75

Variance = Σ (X - μ)² p(X)
               = 0.75² * 1/2 + 0.25² *3/10 + 1.25²*3/20 +2.25²*1/20
               = 0.7875