Probability mass function and distribution for Tossing of Coins

Physics Chemistry  Biology  Mathematics
Science > Mathematics > ProbabilityYou are Here

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 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
Science > Mathematics > ProbabilityYou are Here
Physics Chemistry  Biology  Mathematics

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