The likelihood of an event to occur is known as its probability. The probability distribution is the distribution table formed to obtain the occurrence of each event in the sample space. From the probability distribution table, we can find the mean, standard deviation, and expected value.

**Example 1: **

If a coin is tossed, find the probability distribution of getting tails.

**Solution:**

If a coin is tossed, either 0 tail or 1 tail can be obtained.

Probability of getting zero tail, P(0) = $\frac{1}{2}$

Probability of getting one tail, P(1) = $\frac{1}{2}$

The probability distribution table will be,

If a coin is tossed, find the probability distribution of getting tails.

If a coin is tossed, either 0 tail or 1 tail can be obtained.

Probability of getting zero tail, P(0) = $\frac{1}{2}$

Probability of getting one tail, P(1) = $\frac{1}{2}$

The probability distribution table will be,

X |
0 | 1 |

P(X) | $\frac{1}{2}$ |
$\frac{1}{2}$ |

If two coins are tossed simultaneously, find the probability distribution of getting head.

If two coins are being tossed, there can be either 0 heads, 1 head, or 2 heads.

Sample space, S = {HH, HT, TH, TT}

The probability distribution of getting heads can be shown as:

X |
0 |
1 | 2 |

P(X) | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |

In a class on 100 students, 80 students passed in all subjects, 10 failed in one subject, 7 failed in two subjects and 3 failed in three subjects. Find the probability distribution of the variable for number of subjects a student from the given class has failed in.

Solution:

For a random student,

The probability of failing in 0 subjects, P(X=0) = 0.8

The probability of failing in 1 subjects, P(X=1) = 0.1

The probability of failing in 2 subjects, P(X=2) = 0.07

The probability of failing in 3 subjects, P(X=3) = 0.03

The probability distribution can be shown as:

X | 0 | 1 | 2 | 3 |

P(X) | 0.8 | 0.1 | 0.07 | 0.03 |

A die is having 6 sides has two dots on 3 sides, four dots on 2 sides and six dot on 1 side. Find the probability distribution of getting a number on rolling the die.

Solution:

On rolling the given die, any one of the three numbers, 2, 4, and 6 will come.

Probability of getting 2 dots, P(X = 2) = $\frac{3}{6}$ = $\frac{1}{2}$

Probability of getting 4 dots, P(X = 4) = $\frac{2}{6}$ = $\frac{1}{3}$

Probability of getting 6 dots, P(X = 6) = $\frac{1}{6}$

The probability distribution can be shown as,

X | 2 | 4 | 6 |

P(X) | $\frac{1}{2}$ | $\frac{1}{3}$ | $\frac{1}{6}$ |

Three coins are being tossed, find the probability distribution of getting any number of tails.

Solution:

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

Hence, in tossing three coins number of tails can be 0, 1, 2, 3.

Probability of getting 0 tails, P(X=0) = $\frac{1}{8}$

Probability of getting 1 tails, P(X=1) = $\frac{3}{8}$

Probability of getting 2 tails, P(X=2) = $\frac{3}{8}$

Probability of getting 3 tails, P(X=3) = $\frac{1}{8}$

Probability distribution can be given as:

X | 0 | 1 | 2 |
3 |

P(X) | $\frac{1}{8}$ |
$\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |