Q. 94.1( 20 Votes )

# The random variab

Answer :

Given: A random variable X with its probability distribution.

(a) As we know the sum of all the probabilities in a probability distribution of a random variable must be one. Hence the sum of probabilities of given table:

k + 2k + 3k + 0 = 1

6k = 1 (b) (i) P(X < 2) = ?

P(X < 2) = P(X = 0) + P(X = 1)

= k + 2k

= 3k (ii) P(X ≤ 2) = ?

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= k + 2k + 3k

= 6k (iii) P(X ≥ 2) = ?

P(X ≥ 2) = P(X = 2) + P(X > 2)

= 3k + 0

= 3k Rate this question :

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