Sum
Solve the following problem :
The probability distribution of a discrete r.v. X is as follows.
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| (X = x) | k | 2k | 3k | 4k | 5k | 6k |
Find P(X ≤ 4), P(2 < X < 4), P(X ≤ 3).
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Solution
a. P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 3k + 4k
= 10k
= `(10)/(21)`
b. P(2 < X < 4) = P(X = 3) = 3k = `(3)/(21) = (1)/(7)`
c. P(X ≥ 3) = 1 – P(X < 3)
= 1 – [P(X = 1) + P(X = 2)]
= 1 – (k + 2k) = 1 – 3k
= `1 - (3)/(21)`
= `1 - (1)/(7)`
=`(6)/(7)`.
Is there an error in this question or solution?
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