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Question
The discrete random variable X has the probability function
| X | 1 | 2 | 3 | 4 |
| P(X = x) | k | 2k | 3k | 4k |
Show that k = 0 1
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Solution
P(x = 1) = k
p(x = 2) = 2k
p(x = 3) = 3k
P(x = 4) = 4k
Since P(X = x) is a probability Mass function
`sum_(x = 1)^4` P(X = x) = 1
`sum_("i" = 1)^oo` P(xi) = 1
p(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1
P(x = 1) + P(x = = 2) + P(x = 3) + P(x = 4) = 1
k + 2k + 3k + 4k = 1
10k = 1
k = `1/10`
∴ k = 0.1
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