1.

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»

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

X23456
P(X)2/304/306/308/3010/30

Required mean = E(x) = ∑pixi = 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)



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