Chapter 13: Probability

We revisit the concept of probability with advanced topics like conditional probability, Bayes’ theorem, multiplication theorem, and independence of events.

Conditional Probability

The probability of an event $E$, given that the event $F$ has already occurred, is called conditional probability of $E$ given $F$, denoted by $P(E|F)$. $$ P(E|F) = \frac{P(E \cap F)}{P(F)} $$ where $P(F) \neq 0$.

Multiplication Theorem on Probability

Let $E$ and $F$ be two events associated with a sample space $S$. The probability of simultaneous occurrence of the events $E$ and $F$ is: $$ P(E \cap F) = P(E) P(F|E) = P(F) P(E|F) $$ where $P(E) \neq 0$ and $P(F) \neq 0$.

Independent Events

Two events $E$ and $F$ are said to be independent if the probability of occurrence of one of them is not affected by the occurrence of the other. Mathematically, $E$ and $F$ are independent if: $$ P(E \cap F) = P(E) \cdot P(F) $$

Theorem of Total Probability

Let $\{E_1, E_2, \dots, E_n\}$ be a partition of the sample space $S$, and suppose that each of the events $E_1, E_2, \dots, E_n$ has nonzero probability of occurrence. Let $A$ be any event associated with $S$, then: $$ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + \dots + P(E_n)P(A|E_n) = \sum_{j=1}^n P(E_j) P(A|E_j) $$

Bayes’ Theorem

If $E_1, E_2, \dots, E_n$ are mutually exclusive and exhaustive events associated with a sample space, and $A$ is any event of non-zero probability, then: $$ P(E_i | A) = \frac{P(E_i) P(A | E_i)}{\sum_{j=1}^n P(E_j) P(A | E_j)} $$

Competency-Based Questions

1. [Sample Paper 2024] An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Solution:

Let $R_1$ be the event that the first ball is red and $B_1$ be the event that the first ball is black.

$P(R_1) = 5/10 = 1/2$.

$P(B_1) = 5/10 = 1/2$.

Let $R_2$ be the event that the second ball is red.

If $R_1$ occurs, the urn will contain $5+2 = 7$ red balls and 5 black balls. Total balls = 12.

$P(R_2 | R_1) = 7/12$.

If $B_1$ occurs, the urn will contain 5 red balls and $5+2 = 7$ black balls. Total balls = 12.

$P(R_2 | B_1) = 5/12$.

By Total Probability Theorem:

$P(R_2) = P(R_1)P(R_2 | R_1) + P(B_1)P(R_2 | B_1)$

$P(R_2) = (1/2)(7/12) + (1/2)(5/12)$

$P(R_2) = 7/24 + 5/24 = 12/24 = 1/2$.

2. [CBSE 2021] A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution:

Let $E_1$ be the event of choosing the first bag and $E_2$ be the event of choosing the second bag.

$P(E_1) = 1/2$. $P(E_2) = 1/2$.

Let $A$ be the event of drawing a red ball.

$P(A|E_1) = P(\text{drawing a red ball from Bag 1}) = 4/8 = 1/2$.

$P(A|E_2) = P(\text{drawing a red ball from Bag 2}) = 2/8 = 1/4$.

We need to find $P(E_1|A)$. By Bayes’ Theorem:

$$ P(E_1|A) = \frac{P(E_1) P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} $$

$$ P(E_1|A) = \frac{(1/2)(1/2)}{(1/2)(1/2) + (1/2)(1/4)} $$

$$ P(E_1|A) = \frac{1/4}{1/4 + 1/8} = \frac{1/4}{3/8} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3} $$

3. [CBSE 2019] If A and B are two events such that $P(A) = \frac{1}{4}, P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{8}$. Find $P(\text{not } A \text{ and not } B)$.

Solution:

$P(\text{not } A \text{ and not } B) = P(A’ \cap B’)$.

By De Morgan’s Law, $A’ \cap B’ = (A \cup B)’$.

So, $P(A’ \cap B’) = P((A \cup B)’) = 1 - P(A \cup B)$.

We know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

$P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$.

Therefore, $P(A’ \cap B’) = 1 - \frac{5}{8} = \frac{3}{8}$.