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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) = 0^{2}/5 = 0/5 = 0

∴ P(X = 1) = 1^{2}/5 = 1/5

∴ P(X = 2) = 2^{2}/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 > Probability > You are Here |

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