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Question
A random variable X has the following probability mass function.
| x | 1 | 2 | 3 | 4 | 5 |
| F(x) | k2 | 2k2 | 3k2 | 2k | 3k |
Find P(2 ≤ X < 5)
Solution
P(2 ≤ X < 5) = P(X = 2) + P(X = 3) + P(X = 4)
= `2/36 + 3/36 + 2/6`
= `(2 + 3 + 12)/36 + 17/36`
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