Understanding ‘Probability’- For grade X
Tags: Education Reform, K12, Higher Education, Edtech, Mathematics

In reality, the outcome of any random experiment is uncertain.

Probability is the science of measuring numerically the likeliness of occurrences of an event or to measure the degree (chance) of the possibility of the occurrence of an event.

There are three different approaches to probability:

Experimental/Empirical/Statistical

Theoretical/ Classical

Axiomatic

We decided on using a particular approach based on the nature of the experiment to which the possibility of occurrence of a particular event, has to be measured.

Experimental approach:

In an experimental approach, you perform a random experiment a number of times called trials. The outcomes of the trials are recorded and probabilities can be calculated.

Where, P(E) = No of trials in which the event happened/ The total number of trials performed

Example:

A coin is tossed 500 times with the following frequencies.

Head = 245

Tail=255

Compute the probabilities for each event.

Solution:

P(H) → Probability of getting ahead; P(T) → Probability of getting a Tail

P(H) = No of trials in which the event happened/ The total number of trials performed

P(H) = 245/500;

Similarly, P(T)=255/500

(Note: P(E) = P(H)+P(T) = will always be 1)

Theoretical approach:

In this concept, the important assumption is that all outcomes are equally likely.

Equally likely — The events under consideration have a similar theoretical probability like in the mentioned below

While tossing a coin

P(Tail)

1/2

P(Head)

1/2

Total = 1

By definition,

Let’s say, Probability of occurrence of an event as ‘P(E)’

Thus, P(E) = Number of chances of occurrence/ Total possible outcomes

The total probability will always be one.

P(E) = 1 (100% possibility of occurrence of the event)

Consider, Probability of non-occurrence of an event as P(Ē)

Thus, P(Ē) = 1- P(E)

Thus the chances of occurrence of the event can vary from 0 to 1.

0 ≤ P(E) ≤ 1

Probability Line

Some of the most common examples of random experiments are

Tossing a coin

Rolling a dice

Drawing a card from a pack of cards

Random Experiment: Example 1 — Tossing a coin

The possible outcomes of tossing a coin are,

Head

2. Tails

The Total possible outcome is 2

While tossing a coin there are equal chances of the outcome being ahead or a Tail

% 50 chance of occurrence of the Head. 50% chance of occurrence of Tail

By the formula,

Probability of getting heads = 1/2

Probability of getting tails = 1/2

Random Experiment: Example 2 — Rolling a die

The possible outcomes of rolling a die will be 1,2,3,4,5,6

Total possible outcomes = 6

Probability of getting 3 = 1/6

Similarly, the Probability of getting 5 = 1/6

Random Experiment: Example 3 — Drawing a card from a pack of cards

In a pack of cards, there are a total of 52 cards

There are four suits ( Clubs, Diamonds, Hearts, and Spades)

Each suite has 13 cards

Total possible outcomes = 52

Probability of getting a spade = 13/52

Practice question 1:

Two players A and B are playing a badminton match. It is known that the probability of B winning the match is 0.6. What is the probability of A winning the match?

Solution:

P(A) → Probability of A winning; P(B) → Probability of B winning

We know that → the Probability of B winning, P(B) = 0.6

Total Probability, P(E) = P(A) + P(B) = 1

P(A) = 1- 0.6

P(A) → Probability of winning the match is 0.4

Practice question 2:

A bag contains 3 blue balls, 2 white balls, and 1 red ball. You are picking a ball.

What is the probability that the ball picked up is

i) Red ii) White iii)Blue

Solution:

The total number of balls in the bag are 3blue+2white+1red = 6

Thus, Total possible outcome = 6

P(B) → Probability of choosing a blue ball; P(W) → Probability of choosing a white ball

P(R) → Probability of choosing a red ball

i) To find: P(R)

Number of possibilities of choosing a red ball = 1

P(R)= Number of possibilities of choosing a red ball/ Total number of outcomes

P(R) = ⅙

ii) To find: P(W)

Number of possibilities of choosing a white ball = 2

P(R)= Number of possibilities of choosing a white ball/ Total number of outcomes

P(R) = 2/6

iii) To find: P(B)

Number of possibilities of choosing a blue ball = 3

P(R)= Number of possibilities of choosing a blue ball/ Total number of outcomes

P(R) = 3/6

(Note: Total probability P(E) = P(B)+P(W)+P(R) = 1/6+2/6+3/6 = 6/6 = 1)

Practice question 3:

A carton consists of 100 shirts of which 88 shirts are good, 8 have minor defects and 4 have major defects. John a trader, will only accept the shirt which is good, but Martin, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that-

i) it is acceptable to John?

ii) is it acceptable to Martin?

Solution:

P(J) → Probability of shirts accepted by John

P(M) → Probability of shirts accepted by Martin

P(J) = 88/100 (John only accepts shirts that are good)

P(M) = 96/100 ( Martin only rejects shirts with major defects = 88+8 = 96)

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