Sample spaces having equally likely outcomes

Sample spaces having equally likely outcomes.

If we assume that all outcomes of an experiment are equally likely to occur, then the probability of any event E equals the proportion of outcomes in the sample space that are contained in E.

P (E) = number of points in E / number of points in S

TYPE 1(a) P(A) = n(A)/n(S)

Example 1.1.1

Find the probability that exactly one head appears in a single throw of a fair coin.

Solution:

Formula P(A) = n(A)/n(S)

Here A be the event of getting exactly one head in a single throw of a fair coin.

S --> Sample space = {H, T}

n(S) = 2

n(A) = 1

.'. P(A) = n(A)/n(S) = 1/2

Example 1.1.2

If two dice are rolled, what is the probability that the sum of the upturned faces will be equal to 7?

Solution :

The number of total outcomes is n(S) = 36

Let A = {Sum of the upturned faces will equal 7} = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

n(A) = 6

It is assumed that all the 36 possible outcomes are equally likely.

.'. P(A) = n(A)/n(S) = 6/36 = 1/6

Example 1.1.3

From a pack of 52 cards two cards are drawn the first being replaced before the second is drawn. Find the probability that the first one is a diamond and second is a king.

Solution: 

Let A be an event of drawing a diamond.

P (A) = n(A)/n(S) = 13/52 = 1/4

Let B be an event of drawing a king.

P(B) = n(B)/n(S) = 4/52 = 1/13

A and B are independent events.

Hence P (A ∩ B) = P(A).P(B) = (1/4) (1/13) = 1/52

Example 1.1.4

A bag contains 5 white and 10 red balls. Three balls are taken out at random. Find the probability that all the three balls drawn red.

Solution : Total number of balls = 15

S = {Three balls are taken out of 15}

Number of red balls = 10

A = {Three balls which are red}

Example 1.1.5

If 3 balls are "randomly drawn" from a bowl containing 6 white and 5 black balls, what is the probability that one of the drawn balls is white and the other two black ?

Solution :

The total number of balls = 11

Among this 3 balls are randomly selected from the bowl.

.'. Total number of possible outcomes = n(S) = 11C₃ = 11.10.9 / 1.2.3 = 165

The number of favourable outcomes

(i.e., 1 is white and the other two are black) = n(A) = 6C₁ x 5C₂

= 6 x (5.4/1.2) = 6 x 10 = 60

Hence the required probability P(A) = n(A)/n(S) = 60/165 = 4/11

Example 1.1.6

A lot of integrated circuit chips consists of 10 good, 4 with minor defects and 2 with major defects. Two chips are randomly chosen from the lot. What is the probability that atleast one chip is good? [A.U M/J 2017]

Solution :

Example 1.1.7

A committee of 5 persons is to be selected randomly from a group of 5 men and 10 women. (a) Find the probability that the committee consists of 2 men and 3 women. (b) Find the probability that the committee consists of all women.

Solution :

(a) The number of total outcomes is given by n(S) = 15C₅

It is assumed that "random selection" means that each of the outcomes is equally likely.

Let A = {The committee consists of 2 men and 3 women}

Then n(A) = (5C₂) (10C₃)

(b) Let B = {the event that the committee consists of all women}

n(B) = (5C₀) (10C₅)

Example 1.1.8

Four persons are chosen at random from a group containing 3 men, 2 women and 4 children. Show that the chance that exactly two of them will be children is 10/21. [A.U N/D 2006]

Solution :

Total number of persons = 9

4 persons can be chosen out of 9 persons 9C₄ ways = 9.8.7.6 / 1.2.3.4 = 126 ways

The number of ways of choosing 2 children out of 4 children = 4C₂ ways = 4.3 / 1.2 = 6 ways

The remaining two persons can be choosen from 5 persons (3 men + 2 women) = 5C₂ ways

= 5.4 / 1.2 = 10 ways

.'. The number of favourable case = 4C₂ × 5C₂ ways = 6 × 10 ways = 60 ways.

Required probability = 60/126 = 10/21

Example 1.1.9

Four persons are chosen at random from a group containing 3 men, 2 women and 4 children. Show that the chance that exactly two of them will be children is 10/21. [A.U N/D 2006]

Solution :

Total number of persons = 9

4 persons can be chosen out of 9 persons = 9C₄ ways = 9.8.7.6 / 1.2.3.4 = 126 ways

The number of ways of choosing 2 children out of 4 children = 4C₂ ways = 4.3 / 1.2 = 6 ways

The remaining two persons can be choosen from 5 persons (3 men + 2 women) = 5C₂ ways = 5.4 / 1.2 = 10 ways

.'. The number of favourable case = 4C₂ × 5C₂ ways = 6 × 10 ways = 60 ways.

.'. Required probability = 60/126 = 10/21

Example 1.1.10

Two dice are thrown together. Find the probability that (a) the total of the numbers on the top face is 9 and (b) the top faces are same. [A.U. M/J 2006]

Solution :

(a) Let A be the event which gives the sum of the top numbers as 9.

Favourable cases which gives the total as 9 are (3, 6), (4, 5), (5, 4), (6, 3)

P(A) = 4/36 = 1/9

(b) Let B be the event which gives the top faces are same.

Favourable cases are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)

P(B) = 6/36 = 1/6

Example 1.1.11

Out of (2n + 1) tickets consecutively numbered three are drawn at random. Find the probability that the numbers on them are in arithmetic progression. [A.U. M/J 2006]

Solution:

Out of (2n + 1) tickets, 3 tickets can be drawn in (2n+1)C₃ ways.

.'. total number of exhaustive cases

To find the favourable number of cases, we give all possibilities in which the numbers on the drawn tickets are in A.P. with common difference d = 1, 2, 3, ..., n-1, n (say)

If d = 1, the possibilities are

Similary, if d = n 1, then the possible cases are

If d= n, then there is only one case as (1, n + 1, 2n + 1)

.'. the total number of favourable cases is (2n-1) + (2n-3) + ... + 5 + 3+1 which is an A.P. with common difference 2.

Hence favourable cases = 

Thus the required probability =