Science > Mathematics > Statistics and Probability > Probability > Probability Distribution
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.