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



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