Q. 245.0( 2 Votes )
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.
Answer :
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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