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State if the following is not the probability mass function of a random variable. Give reasons for your answer

Z |
3 | 2 | 1 | 0 | −1 |

P(Z) |
0.3 | 0.2 | 0.4 | 0 | 0.05 |

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#### Solution 1

**P.m.f. of random variable should satisfy the following conditions:**

- 0 ≤ p
_{i}≤ 1 - ∑p
_{i}= 1

Z |
3 | 2 | 1 | 0 | −1 |

P(Z) |
0.3 | 0.2 | 0.4 | 0 | 0.05 |

Here ∑p_{i} = 0.3 + 0.2 + 0.4 + 0 + 0.05

= 0.95 ≠ 1

Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

#### Solution 2

Here, p_{i} ≥ 0, `AA` i = 1, 2, ...., 5

Now consider,

`sum_("i" = 1)^5 "P"_"i"` = 0.3 + 0.2 + 0.4 + 0 + 0.05

= 0.95 ≠ 1

∴ Given distribution is not p.m.f.

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