Physics |
Chemistry |
Biology |
Mathematics |

Science > Mathematics > Probability > You 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 = 2^{4} = 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) = ^{4}C_{0} /16 = 1/16

P( X = 1) = P(1) = ^{4}C_{1} /16 = 4/16 = 1/4

P( X = 2) = P(2) = ^{4}C_{2} /16 = 6/16 = 3/8

P( X = 3) = P(3) = ^{4}C_{3} /16 = 4/16 = 1/4

P( X = 4) = P(4) = ^{4}C_{4} /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 = 2^{5} = 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) = ^{5}C_{0} /32 = 1/32

P( X = 1) = P(1) = ^{5}C_{1} /32 = 5/32

P( X = 2) = P(2) = ^{5}C_{2} /32 = 10/32 = 5/16

P( X = 3) = P(3) = ^{5}C_{3} /32 = 10/32 = 5/16

P( X = 4) = P(4) = ^{5}C_{4} /32 = 5/32

P( X = 5) = P(5) = ^{5}C_{5} /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 = 2^{6} = 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) = ^{6}C_{0} /64 = 1/64

P( X = 1) = P(1) = ^{6}C_{1} /64 = 6/64 = 3/32

P( X = 2) = P(2) = ^{6}C_{2} /64 = 15/64

P( X = 3) = P(3) = ^{6}C_{3} /64 = 20/64 = 5/16

P( X = 4) = P(4) = ^{6}C_{4} /64 = 15/64

P( X = 5) = P(5) = ^{6}C_{5} /64 = 6/64 = 3/32

P( X = 6) = P(6) = ^{6}C_{6} /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 > Probability > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |