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Question
Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by
`f(x) = {{:((x^2 + 1)/k"," "for" x = 0"," 1"," 2),(0"," "otherwise"):}`
Find P(X ≥ 1)
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Solution
P(X ≥ 1) = = P(X = 1) + P(X = 2)
= `2/8 + 5/8 + 7/8`
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