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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})}

Here, P(E) is the Probability or likelihood of an outcome, n\bm{n}(E) is the number of favourable outcomes and n\bm{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}.
Number of possible outcomes =n(S)=6= n\text{(S)} = 6

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

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

Answer:12\dfrac{1}{2}

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

11) 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.

22) Sample Space: This is the set of all possible outcomes. In the example above, the sample space for rolling a six-sided die contain 66 outcomes −- {1,2,3,4,5,61, 2, 3, 4, 5, 6}.

33) 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 33 outcomes −- {2,3,52, 3, 5}.

44) 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)}}

2. Basic Properties

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

Probability will always be between 0\bm{0} and 1\bm{1} (both inclusive). The number of favourable outcomes is between 00 and total possible outcomes, i.e. 0≤0 \leq 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 \leq P(E) ≤1\leq1

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

The probability of an event, P(E), subtracted from 11 gives the probability for the non-occurrence of that event, P(E‾)(\overline{\text{E}}). Where the number of favourable outcomes is nn(E), the number of unfavourable outcomes is nn(S) −- nn(E).

P(E‾)\overline{\text{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})

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

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

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

Example 2

A pack of 8080 cards contains 55 suits. Each suit contains differently numbered cards with first 1616 natural numbers. When a card is randomly drawn, what is the probability of the card's number being more than 55?

Solution

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

There are 1111 numbers greater than 55 and less than or equal to 1616.
Favourable outcomes =11×5=55= 11 \times 5 = 55

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

Answer: 1116\dfrac{11}{16}

Example 3

From the set of all 44-digit numbers where no digit is repeated, a number is randomly selected. What is the probability of this 44-digit number being even?

Solution

Number of 44-digit numbers without repetition =n(S)=9×9×8×7= n(\text{S}) = 9 \times 9 \times 8 \times 7

As the favourable outcomes are those which are even numbers, their units digit would be 0,2,4,60, 2, 4, 6 or 88. As 00 is involved, we look at it as two cases

Case I: When units digit is 00,
Number of 44-digit numbers =9×8×7×1= 9 \times 8 \times 7 \times 1

Case II: When units digit is 2,4,6or82, 4, 6 \mathrm{or} 8,
Number of 44-digit numbers =8×8×7×4= 8 \times 8 \times 7 \times 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)
=8×7(9+32)= 8 \times 7(9 + 32)
=41×8×7= 41 \times 8 \times 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}

Answer: 4181\dfrac{41}{81}

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