### Probability - Basic Concepts

• Understand basic terminology related to probability such as sample space, event, complement, boolean combination and dependency of multiple events.

An event is a set of outcomes of an experiment. Sample space is defined as set of all possible outcomes.
Probability of an event is defined as ratio of number of favourable outcomes to the number of outcomes in the sample space.
Probabilities can be expressed as proportions that range from 0 to 1, and they can also be expressed as percentages ranging from 0% to 100%.

• Tossing a coin. The sample space is S = H,T.

• Tossing a die. The sample space is S = 1,2,3,4,5,6.

• Tossing a coin twice. The sample space is S = HH,HT,TH,TT.

• Tossing a die twice. The sample space is S = (i,j) : i,j = 1,2,...,6, which contains 36 elements.

Conditional probability is a measure of the probability of an event given that another event has already occurred.

Intersection of two events is denoted by A $\cap$ B and is shown in the following Venn Diagram.
Union of two events is denoted by A $\cup$ B and is shown in the following Venn Diagram.
In Permutations, order matters.

$P(n,r) = \dfrac{n!}{n-r!}$ In Combinations, order does not matter.

$C(n,r) = \dfrac{n!}{r!(n-r)!}$

### Solved Example:

#### 7-1-01

From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be Kings, if first card in NOT replaced?

Solution:
Given : Total number of cards = 52 and two cards are drawn at random. Number of kings in playing cards = 4 So the probability that both cards will be king is given by, $P = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{221}$

### Solved Example:

#### 7-1-02

Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight?

Solution:
We know that a dice has 6 faces and 6 numbers so the total number of cases (outcomes) = 6$\times$6 = 36 And total ways in which sum of the numbers on the dices is eight, $\left\{(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)\right\}$
So, the probability that the sum of the numbers eight is, p =$\dfrac{5}{36}$

### Solved Example:

#### 7-1-03

A box contains 5 black and 5 red balls. Two balls are randomly picked one after another form the box, without replacement. The probability for balls being red is:

Solution:
$\mathrm{Black\ balls} = 5, \mathrm{Red\ balls} = 5, \mathrm{Total\ balls} = 10$ Here, two balls are picked from the box randomly one after the other without replacement. So the probability of both the balls are red is: $P =\dfrac {5C_{0}\times 5C_{2}}{10C_{2}} =\dfrac {\dfrac {5!}{0!5!}\times \dfrac {5!}{3!2!}}{\dfrac {10!}{3!2!}} =\dfrac {1\times 10}{45} =\dfrac {2}{9}$
Alternate Method: The probability of drawing a red ball, $P_1 = \dfrac{5}{10} = \dfrac{1}{2}$
If ball is not replaced, then box contains 9 balls. So, probability of drawing the next red ball from the box. $P_2= \dfrac{4}{9}$ Hence, probability for both the balls being red is, $P = P_1 \times P_2 = \dfrac{1}{2} \times \dfrac{4}{9} = \dfrac{2}{9}$

### Solved Example:

#### 7-1-04

A single die is thrown twice. What is the probability that the sum is neither 8 nor 9?

Solution:
We know a die has 6 faces and 6 numbers so the total number of ways = 6$\times$6 = 36. And total ways in which sum is either 8 or 9 is 9, i.e. $(2,6), (3,6) (3,5) (4,4) (4,5) (5,4) (5,3) (6,2) (6,3)$
Total number of tosses when both the 8 or 9 numbers are not come = 36 - 9 = 27. Then probability of not coming sum 8 or 9 is, 36 = $\dfrac{27}{36}$ = $\dfrac{3}{4}$

### Solved Example:

#### 7-1-05

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ?

Solution:
Sample Space= Any two items can be selected from 100 items = $100_{C_2}$
Favourable case= Defective two items can be selected from 100 items = $20_{C_2}$ Probability of the required event is given by: $= \dfrac{20_{C_2}}{100_{C_2}} = \dfrac{\dfrac{20!}{18!2!}}{\dfrac{100!}{98!2!}} = \dfrac{20 \times 19}{100 \times 99} = \dfrac{19}{495}$

### Solved Example:

#### 7-1-06

A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is:

Solution:
The box contains:
Number of washers = 2, Number of nuts = 3, Number of bolts = 4
Total objects = 2 + 3 + 4 = 9
First two washers are drawn from the box which contain 9 items. So the probability of drawing 2 washers is, $P_{1} =\dfrac {2_{C_{2}}}{9_{C_{2}}} =\dfrac {1}{\left( \dfrac {9!}{7!2!}\right)} =\dfrac {7!2!}{9\times 8\times 7!} =\dfrac {2}{9\times 8} =\dfrac{1}{36}$ After this box contains only 7 objects and then 3 nuts drawn from it. So the probability of drawing 3 nuts from the remaining objects is, $P_{2} =\dfrac {3_{C_{3}}}{7_{C_{3}}} =\dfrac {1}{\left( \dfrac {7!}{4!3!}\right) } =\dfrac {4!3!}{7\times 6\times 5\times 4!} =\dfrac {1}{35}$ After this box contain only 4 objects, probability of drawing 4 bolts from the box, $P_{3}=\dfrac {4_{C_{4}}}{4_{C_{4}}} =\dfrac {1}{1} =1$
Therefore the required probability is, $P = P_1P_2P_3 = \dfrac{1}{36} \times \dfrac{1}{35} \times 1 = \dfrac{1}{1260}$

### Solved Example:

#### 7-1-07

A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is:

Solution:
No. of Red balls = 4, No. of Black ball = 6
3 balls are selected randomly one after another, without replacement.
1 red and 2 black balls are will be selected as following
$R B B: \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{1}{6},$ $B R B: \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} = \dfrac{1}{6},$ $B B R: \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{1}{6}$ Hence Total probability of selecting 1 red and 2 black ball is: $P = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2}$