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The probability distribution of X is as follows: x 0 1 2 3 4 P[X = x] 0.1 k 2k 2k k Find k P[X < 2] P[X ≥ 3] P[1 ≤ X < 4] P(2) - Mathematics and Statistics

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Question

The probability distribution of X is as follows:

x 0 1 2 3 4
P[X = x] 0.1 k 2k 2k k

Find

  1. k
  2. P[X < 2]
  3. P[X ≥ 3]
  4. P[1 ≤ X < 4]
  5. P(2)
Sum

Solution

a. The table gives a probability distribution and therefore P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] = 1

i.e., 0.1 + k + 2k + 2k + k = 1

i.e., 6k = 0.9

∴ k = 0.15

k = 0.15

b. P[X < 2] = P[X = 0] + P[X = 1] = 0.1 + k

= 0.1 + 0.15

= 0.25

c. P[X ≥ 3] = P[X = 3] + P[X = 4] = 2k + k

= 3k

= 3(0.15)

= 0.45

d. P[1 ≤ X < 4] = P[X = 1] + P[X = 2] + P[X = 3]

= k + 2k + 2k

= 5k

= 5(0.15)

= 0.75

e. P(2) = P[X ≤ 2] = P[X = 0] + P[X = 1] + P[X = 2]

= 0.1 + k + 2k

= 0.1 + 3k

= 0.1 + 0.45

= 0.55

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Probability Distribution of Discrete Random Variables
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