In this article, we shall study to check whether the given function is a probability mass function or not.

Example – 01: 

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

Solution:

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: 

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

Solution:

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: 

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

Solution:

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: 

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

Solution:

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: 

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

Solution:

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: 

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

Solution:

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: 

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

Solution:

∴   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

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: 

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

Solution:

∴   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

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

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

Solution:

∴   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

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.

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In this article, we shall study to write probability mass function and to write probability distribution for the given event.

Example – 01:

If a coin is tossed two times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed two times. The sample space for the experiment is as follows

S = {HH, HT, TH, TT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails)

Probability mass function is

P( X = 0) = P(0) = 1/4

P( X = 1) = P(1) = 2/4 = 1/2

P( X = 2) = P(02) = 1/4
Hence, The probability distribution for the number of tails is as follows.

Example – 02:

If a coin is tossed three times and X denotes the number of tails. Find the probability mass function of X. Also write the probability distribution of X.
Solution:

If a coin is tossed three times. The sample space for the experiment is as follows

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8

The probability distribution for the number of tails is as follows.

Example – 03:

If a coin is tossed four times and X denotes the number of tails. Find the probability distribution of X.

Solution:

METHOD- I:

If a coin is tossed four times. The sample space for the experiment is as follows

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, TTHT, TTTH, TTTT}

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = 1/8
P( X = 1) = P(1) = 3/8
P( X = 2) = P(2) = 3/8
P( X = 3) = P(3) = 1/8
P( X = 4) = P(4) = 3/8

The probability distribution for the number of tails is as follows.

METHOD- II

If a coin is tossed four times. The sample space for the experiment is as follows

S = 2⁴ = 16

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁴C₀ /16 = 1/16
P( X = 1) = P(1) = ⁴C₁ /16 = 4/16 = 1/4
P( X = 2) = P(2) = ⁴C₂ /16 = 6/16 = 3/8
P( X = 3) = P(3) = ⁴C₃ /16 = 4/16 = 1/4
P( X = 4) = P(4) = ⁴C₄ /16 = 1/16

The probability distribution for the number of tails is as follows.

Example – 04:

If a coin is tossed five times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed five times. The sample space for the experiment is as follows

S = 2⁵ = 32

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁵C₀ /32 = 1/32
P( X = 1) = P(1) = ⁵C₁ /32 = 5/32
P( X = 2) = P(2) = ⁵C₂ /32 = 10/32 = 5/16
P( X = 3) = P(3) = ⁵C₃ /32 = 10/32 = 5/16
P( X = 4) = P(4) = ⁵C₄ /32 = 5/32
P( X = 5) = P(5) = ⁵C₅ /32 = 1/32

The probability distribution for the number of tails is as follows.

Example – 05:

If a coin is tossed six times and X denotes the number of tails. Find the probability distribution of X.

Solution:

If a coin is tossed six times. The sample space for the experiment is as follows

S = 2⁶ = 64

Given that X denotes the number of tails. X can take values 0 (No tail) or 1 (One tail) or 2 (two tails) or 3 (three tails) or 4 (four tails) or 5 (five tails) or 6 (six tails)

Hence the probability mass function is given by

P( X = 0) = P(0) = ⁶C₀ /64 = 1/64
P( X = 1) = P(1) = ⁶C₁ /64 = 6/64 = 3/32
P( X = 2) = P(2) = ⁶C₂ /64 = 15/64
P( X = 3) = P(3) = ⁶C₃ /64 = 20/64 = 5/16
P( X = 4) = P(4) = ⁶C₄ /64 = 15/64
P( X = 5) = P(5) = ⁶C₅ /64 = 6/64 = 3/32
P( X = 6) = P(6) = ⁶C₆ /64 = 1/64

The probability distribution for the number of tails is as follows.

In the next article, we shall study problems in which we will be checking, whether the distribution given is probability mass function or not.

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