Question

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution. 

Answer 1

First six positive integers are 1, 2, 3, 4, 5, 6 

If two numbers are selected at random from above six numbers then sample space S is given by 

S = {(1, 2)(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4),(2, 5), (2, 6), (3, 1),(3, 2), (3, 4),(3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

n(s) = 30. 

Here, X is random variable, which may have value 2, 3, 4, 5 or 6. 

Therefore, required probability distribution is given as 

P(X = 2) = Probability of event getting (1, 2), (2, 1) = 2/30 

P(X = 3) = Probability of event getting (1, 3), (2, 3), (3, 1), (3, 2) = 4/30 

P(X = 4) = Probability of event getting (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3) = 6/30 

P(X = 5) = Probability of event getting (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4) = 8/30 

P(X = 6) = Probability of event getting (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) = 10/30 

It is represented in tabular form as

X	2	3	4	5	6
P(X)	2/30	4/30	6/30	8/30	10/30

Required mean = E(x) = ∑p_ix_i = 2 x 2/30 + 3 x 4/30 + 4 x 6/30 + 5 x 8/30 + 6 x 10/30

= (4 + 12 + 24 + 40 + 60)/30 = 140/30 = 14/3 = 4(2/3)