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Probability

Probability

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Basics of Probability
Mutually Exclusive Events
Non-Mutually Exclusive Events
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Probability : Level 1
Probability : Level 2
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CAT 2025 Lesson : Probability - Basics of Probability

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1. Introduction

Probability is a measure of the likelihood of the occurrence of an event. It is the number of favourable outcomes divided by the total number of outcomes. It does not tell us what will be the outcome, but instead tells us how likely a particular outcome is. Where S is the sample space of all events or outcomes and E is the set of expected/favourable/desired events or outcomes,

P(E)=n(E)n(S) = \dfrac{n(\mathrm{E})}{n(\mathrm{S})}=n(S)n(E)​

Here, P(E) is the Probability or likelihood of an outcome, n\bm{n}n(E) is the number of favourable outcomes and n\bm{n}n(S) is the total possible outcomes.

Example 1

What is the probability of getting a prime number when a six-sided die is rolled?

Solution

Total possible outcomes =S=1,2,3,4,5,6= \text{S} = {1, 2, 3, 4, 5, 6}=S=1,2,3,4,5,6.
Number of possible outcomes =n(S)=6= n\text{(S)} = 6=n(S)=6

Favourable outcomes === Prime numbers =E=2,3,5= \text{E} = {2, 3, 5}=E=2,3,5
Number of favourable outcomes =n(E)=3= n\text{(E)} = 3=n(E)=3

P(E) =n(E)n(S)=36=12 = \dfrac{n\text{(E)}}{n\text{(S)}} = \dfrac{3}{6} = \dfrac{1}{2}=n(S)n(E)​=63​=21​

Answer:12\dfrac{1}{2}21​

A few of the basic terms pertaining to probability are as listed below.

111) Experiment, also called Random Experiment, is one where the result or outcome cannot be predicted. However, it is possible to list all the outcomes and assess the likelihood of each, i.e., find the probability of one or more of the outcomes. In the example above, rolling a six-sided die is an experiment.

222) Sample Space: This is the set of all possible outcomes. In the example above, the sample space for rolling a six-sided die contain 666 outcomes −-− {1,2,3,4,5,61, 2, 3, 4, 5, 61,2,3,4,5,6}.

333) Event: This is the desired outcome or set of outcomes for which we find the probability or likelihood of occurrence. In the example above, the event is getting a prime number when the six-sided die is rolled, i.e. getting one of these 333 outcomes −-− {2,3,52, 3, 52,3,5}.

444) Probability of the Event: P(E) =Number of favourable outcomesTotal possible outcomes=n(E)n(S)= \dfrac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \dfrac{n\text{(E)}}{n\text{(S)}}=Total possible outcomesNumber of favourable outcomes​=n(S)n(E)​

2. Basic Properties

Property 1\bm{1}1: 0≤\bm{0 \leq}0≤ P(E) ≤1\bm{\leq 1}≤1

Probability will always be between 0\bm{0}0 and 1\bm{1}1 (both inclusive). The number of favourable outcomes is between 000 and total possible outcomes, i.e. 0≤0 \leq0≤ n(E)≤n(S)⇒0≤n(E)n(S)≤1⇒0≤n\text{(E)} \leq n\text{(S)} ⇒ 0 \leq \dfrac{n\text{(E)}}{n\text{(S)}} \leq 1 ⇒ 0 \leqn(E)≤n(S)⇒0≤n(S)n(E)​≤1⇒0≤ P(E) ≤1\leq1≤1

Property 2\bm{2}2: P (E) === 1−\bm{1 -}1− P(E)

The probability of an event, P(E), subtracted from 111 gives the probability for the non-occurrence of that event, P(E‾)(\overline{\text{E}})(E). Where the number of favourable outcomes is nnn(E), the number of unfavourable outcomes is nnn(S) −-− nnn(E).

P(E‾)\overline{\text{E}})E) === n(S)−n(E)n(S)=1−n(E)n(S)=1−P(E)\dfrac{n\text{(S)} - n\text{(E)}}{n(\text{S})}=1-\dfrac{n\text{(E)}}{n\text{(S)}}=1 -\text{P}(\text{E})n(S)n(S)−n(E)​=1−n(S)n(E)​=1−P(E)

Property 3\bm{3}3: P(A∪B)\bm{\text{P}(\text{A} \cup \text{B})}P(A∪B) === P(A)+P(B)−P(A∩B)\bm{\text{P}(\text{A})+\text{P}(\text{B}) - \text{P}(\text{A} \cap \text{B})}P(A)+P(B)−P(A∩B)

In the Set Theory lesson we have learnt : n(A∪B)n(\text{A} \cup \text{B})n(A∪B) === n(A)+n(B)−n(A∩B)n(A)+n(B)-n(A \cap B)n(A)+n(B)−n(A∩B)

∴\therefore∴ P(A∪B)\text{P}(\text{A} \cup \text{B})P(A∪B) === n(A∪B)n(\text{A} \cup \text{B})n(A∪B) === n(A∪B)n(S)\dfrac{n(\text{A} \cup \text{B})}{n(\text{S})}n(S)n(A∪B)​ === n(A)+n(B)−n(A∩B)n(S)\dfrac{n(\text{A})+n(\text{B})-n(\text{A} \cap\text{B})}{n(\text{S})} n(S)n(A)+n(B)−n(A∩B)​ === P(A)+P(B)−P(A∩B) \text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A} \cap \text{B})P(A)+P(B)−P(A∩B)

Example 2

A pack of 808080 cards contains 555 suits. Each suit contains differently numbered cards with first 161616 natural numbers. When a card is randomly drawn, what is the probability of the card's number being more than 555?

Solution

Total outcomes =16×5=80= 16 \times 5 = 80=16×5=80

There are 111111 numbers greater than 555 and less than or equal to 161616.
Favourable outcomes =11×5=55= 11 \times 5 = 55=11×5=55

Probability =5580=1116= \dfrac{55}{80} = \dfrac{11}{16}=8055​=1611​

Answer: 1116\dfrac{11}{16}1611​

Example 3

From the set of all 444-digit numbers where no digit is repeated, a number is randomly selected. What is the probability of this 444-digit number being even?

Solution

Number of 444-digit numbers without repetition =n(S)=9×9×8×7= n(\text{S}) = 9 \times 9 \times 8 \times 7=n(S)=9×9×8×7

As the favourable outcomes are those which are even numbers, their units digit would be 0,2,4,60, 2, 4, 60,2,4,6 or 888. As 000 is involved, we look at it as two cases

Case I: When units digit is 000,
Number of 444-digit numbers =9×8×7×1= 9 \times 8 \times 7 \times 1=9×8×7×1

Case II: When units digit is 2,4,6or82, 4, 6 \mathrm{or} 82,4,6or8,
Number of 444-digit numbers =8×8×7×4= 8 \times 8 \times 7 \times 4=8×8×7×4

Total number of favourable outcomes =(9×8×7×1)+(8×8×7×4)= (9 \times 8 \times 7 \times 1) + (8 \times 8 \times 7 \times 4)=(9×8×7×1)+(8×8×7×4)
=8×7(9+32)= 8 \times 7(9 + 32)=8×7(9+32)
=41×8×7= 41 \times 8 \times 7=41×8×7

Probability =41×8×79×9×8×7=4181= \dfrac{41 \times 8 \times 7}{9 \times 9 \times 8 \times 7} = \dfrac{41}{81}=9×9×8×741×8×7​=8141​

Answer: 4181\dfrac{41}{81}8141​

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