Chapter 14: Probability
14.1 Introduction
Probability is the measure of uncertainty of various phenomena. In mathematical terms, probability defines the likelihood of occurrence of an event.
- Random Experiment: An experiment whose outcomes cannot be predicted with certainty.
- Sample Space (S): The set of all possible outcomes of a random experiment.
- Event (E): A subset of the sample space.
14.2 Types of Events
- Impossible Event: The empty set \( \emptyset \). It indicates an event that cannot occur.
- Sure Event: The entire sample space \( S \).
- Simple Event: An event containing only one sample point.
- Compound Event: An event containing more than one sample point.
Algebra of Events
- Complementary Event (Not E): The event ‘\( \text{not } E \)’ occurs when \( E \) does not occur. Denoted by \( E’ \). \( E’ = S - E \).
- Event ‘A or B’: Occurs if either \( A \) occurs, or \( B \) occurs, or both occur. Denoted by \( A \cup B \).
- Event ‘A and B’: Occurs if both \( A \) and \( B \) occur simultaneously. Denoted by \( A \cap B \).
- Mutually Exclusive Events: Two events \( A \) and \( B \) are mutually exclusive if they cannot occur simultaneously. \( A \cap B = \emptyset \).
- Exhaustive Events: Events \( E_1, E_2, \dots, E_n \) are exhaustive if their union equals the sample space. \( E_1 \cup E_2 \cup \dots \cup E_n = S \).
14.3 Axiomatic Approach to Probability
Let \( S \) be the sample space of a random experiment. A probability \( P \) is a real valued function whose domain is the power set of \( S \) and range is the interval \( [0, 1] \) satisfying the following axioms:
- For any event \( E \), \( P(E) \ge 0 \).
- \( P(S) = 1 \).
- If \( E \) and \( F \) are mutually exclusive events, then \( P(E \cup F) = P(E) + P(F) \).
Important Theorems
- \( P(E’) = 1 - P(E) \)
- \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
- If \( A \) and \( B \) are mutually exclusive, \( P(A \cap B) = 0 \), so \( P(A \cup B) = P(A) + P(B) \).
Competency-Based Questions
Question 1:
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that:
(a) Both अनिल and Ashima will not qualify the examination.
(b) At least one of them will not qualify the examination.
Answer 1:
Let \( A \) be the event “Anil qualifies” and \( B \) be the event “Ashima qualifies”.
Given:
\( P(A) = 0.05 \)
\( P(B) = 0.10 \)
\( P(A \cap B) = 0.02 \)
(a) Both will not qualify:
This is the event \( A’ \cap B’ \). By De Morgan’s Laws, \( A’ \cap B’ = (A \cup B)’ \).
First, find \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.05 + 0.10 - 0.02 = 0.13 \]
Now, \( P(A’ \cap B’) = P((A \cup B)’) = 1 - P(A \cup B) = 1 - 0.13 = 0.87 \]
.
(b) At least one of them will not qualify:
This is the event “not both qualify”, which is \( (A \cap B)’ \).
\[ P((A \cap B)’) = 1 - P(A \cap B) = 1 - 0.02 = 0.98 \]
.
Conclusion: The probability that neither qualifies is 0.87, and the probability that at least one does not qualify is 0.98.
Question 2:
A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be:
(i) red
(ii) yellow
(iii) blue
(iv) not blue
(v) either red or blue
Answer 2:
Total number of discs = \( 4 + 3 + 2 = 9 \).
(i) Probability of red:
\( P(R) = \frac{\text{Number of Red Discs}}{\text{Total Discs}} = \frac{4}{9} \)
(ii) Probability of yellow:
\( P(Y) = \frac{2}{9} \)
(iii) Probability of blue:
\( P(B) = \frac{3}{9} = \frac{1}{3} \)
(iv) Probability of not blue:
\( P(B’) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
(v) Probability of either red or blue:
Since a single disc cannot be both red and blue, events R and B are mutually exclusive.
\( P(R \cup B) = P(R) + P(B) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9} \)
Conclusion: The answers are (i) 4/9, (ii) 2/9, (iii) 1/3, (iv) 2/3, and (v) 7/9.