Checking of Probability Mass Function

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  • Verify whether the function can be regarded as probability mass function (p.m.f.)

Example – 01: 

X = x 1 2 3 4
P(X=x) 0.1 0.2 0.3 0.4
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 1 ≤ x ≤ 4) = P(1) + P(2) + P(3) + P(4)

∴   ∑ P(X = x, 1 ≤ x ≤ 4) = 0.1 + 0.2 + ).3 + 0.4 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.



Hence given function is a p.m.f.

Example – 02: 

X = x 0 1 2 3
P(X=x) 0.5 0.2 0.18 0.12
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 3) = P(0) + P(1) + P(2) + P(3)

∴   ∑ P(X = x, 0 ≤ x ≤ 3) = 0.5 + 0.2 + 0.18 + 0.12 = 1



Thus sum of all probabilities is 1. Hence the second condition is satisfied.

Hence given function is a p.m.f.

Example – 03: 

X = x -1 0 1
P(X=x) – 0.2 1 0.2
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.2 < 0 Hence the first condition is not satisfied.

Hence given function is not a p.m.f.

Example – 04: 

X = x 0 1 2
P(X=x) 0.6 0.1 0.2
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.



  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, 0 ≤ x ≤ 1 = P(0) + P(1) + P(2)

∴   ∑ P(X = x, 0 ≤ x ≤ 2) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Thus sum of all probabilities is not equal to1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 05: 

X = x 2 4 6 8
P(X=x) 0.2 0.4 0.6 0.8
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.



  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 2, 4, 6, 8) = P(2) + P(4) + P(6) + P(8)

∴   ∑ P(X = x, x = 2, 4, 6, 8) = 0.2 + 0.4 + 0.6 + 0.8 = 2 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Example – 06: 

X = x -2 -1 1 2
P(X=x) 0.5 -0.1 0.6 0
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that P(-1) = – 0.1 < 0 Hence the first condition is not satisfied.



Hence given function is not a p.m.f.

Example – 07: 

∴   P(X = 0) = 02/5 = 0/5 = 0

∴   P(X = 1) = 12/5 = 1/5



∴   P(X = 2) = 22/5 = 4/5

The probability distribution is

X = x 0 1 2
P(X=x) 0 1/5 4/5
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 0, 1, 2) = P(0) + P(1) + P(2)

∴   ∑ P(X = x, x = 0, 1, 2) = 0 + 1/5 + 4/5 = 5/5 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.



Hence given function is a p.m.f.

Example – 08: 

∴   P(X = 0) = (1- 1)/3 = 0/3 = 0

∴   P(X = 1) = (2 – 1)/3 = 1/3



∴   P(X = 2) = (3 – 1)/3 = 2/3

The probability distribution is

X = x 1 2 3
P(X=x) 0 1/3 2/3
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 1, 2, 3) = P(1) + P(2) + P(3)

∴   ∑ P(X = x, x = 1, 2, 3) = 0 + 1/3 + 2/3 = 1

Thus sum of all probabilities is 1. Hence the second condition is satisfied.



Hence given function is a p.m.f.

Example – 09: 

The function is P( X = x) = (x – 5)/4, x = 5.5, 6.5, 7.5

∴   P(X = 5.5) = (5.5 – 5)/4 = 0.5/4 = 1/8

∴   P(X = 6.5) = (6.5 – 5)/4 = 1.5/4 = 3/8

∴   P(X = 7.5) = (7.5 – 5)/4 = 2.5/4 = 5/8

The probability distribution is

X = x 5.5 6.5 7.5
P(X=x) 1/8 3/8 5/8
  • Checking of the first condition (P(X = x) ≥ 0, ∀ x)

We can see that all the values of P(X = x) ≥ 0. Hence the first condition is satisfied.

  • Checking of the second condition ∑ P(X = x) = 1

∑ P(X = x, x = 5.5, 6.5, 7.5) = P(5.5) + P(6.5) + P(7.5)

∴   ∑ P(X = x, x = 5.5, 6.5, 7.5) = 1/8 + 3/8 + 5/8 = 9/8 ≠ 1

Thus sum of all probabilities is not equal to 1. Hence the second condition is not satisfied.

Hence given function is not a p.m.f.

Science > Mathematics > ProbabilityYou are Here
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