Probability Mass Function and Probability Distribution

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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.

X	0	1	2
P(X)	1/4	1/2	1/4

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.

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

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.

X	0	1	2	3	4
P(X)	1/16	1/4	3/8	1/4	1/16

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.

X	0	1	2	3	4
P(X)	1/16	1/4	3/8	1/4	1/16

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.

X	0	1	2	3	4	5
P(X)	1/32	5/32	5/16	5/16	5/32	1/32

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.

X	0	1	2	3	4	5	6
P(X)	1/64	3/32	15/64	5/16	15/64	3/32	1/64

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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