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Question
A random variable X has the following probability distribution.
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | k2 |
2k2 |
7k2 + k |
Determine
(i) k
(ii) P (X < 3)
(iii) P (X > 6)
(iv) P (0 < X < 3)
Solution
(i) It is known that the sum of probabilities of a probability distribution of random variables is one.
k = − 1 is not possible as the probability of an event is never negative.
`:.k = 1/10`
(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)
(iii) P (X > 6) = P (X = 7)
(iv) P (0 < X < 3) = P (X = 1) + P (X = 2)
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