Answer :

Given: A random variable X with its probability distribution.

(i) 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:

⇒ 0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7K^{2} + k = 1

⇒ 10K^{2} + 9k = 1

⇒ 10K^{2} + 9k – 1 = 0

⇒ (10K-1)(k + 1) = 0

It is known that probability of any observation must always be positive that it can’t be negative.

So

(ii) P(X < 3) = ?

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k

(iii) P(X > 6) = ?

P(X > 6) = P(X = 7)

= 7K^{2} + k

(iv) P(0 < X < 3) = ?

P(0 < X < 3) = P(X = 1) + P(X = 2)

= k + 2k

= 3k

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