Q24 of 26 Page 1

Three numbers are selected at random (without replacement) from first six positive integers. Let X denote the largest of the three numbers obtained. Find the probability distribution of X. Also, find the mean and variance of the distribution.

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


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