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Q4:
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Solution :

Step 1: Defining the Sample Space ($S$)

A fair coin has two outcomes: Heads ($H$) and Tails ($T$). An unbiased six-sided die has outcomes $\{1, 2, 3, 4, 5, 6\}$. The sample space $S$ for the combined experiment is the Cartesian product of these sets:

$S = \{ (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) \}$

The total number of outcomes is $n(S) = 2 \times 6 = 12$. [By the Fundamental Counting Principle]

Step 2: Defining Events A and B

Event $A$: 'Head appears on the coin'

$A = \{ (H,1), (H,2), (H,3), (H,4), (H,5), (H,6) \}$

$n(A) = 6$

$P(A) = \frac{n(A)}{n(S)} = \frac{6}{12} = \frac{1}{2}$

Event $B$: '3 on the die'

$B = \{ (H,3), (T,3) \}$

$n(B) = 2$

$P(B) = \frac{n(B)}{n(S)} = \frac{2}{12} = \frac{1}{6}$

Step 3: Calculating the Intersection ($A \cap B$)

The intersection event $A \cap B$ represents the occurrence of both 'Head' and '3'.

$A \cap B = \{ (H,3) \}$

$n(A \cap B) = 1$

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{12}$

Step 4: Evaluating Independence

Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \times P(B)$. [Per the Definition of Independent Events]

Calculating the product $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Visual Representation of Sample Space

Sample Space S (12 outcomes) Set A (Head) Set B (3) A ∩ B = {(H,3)}

Conclusion

Since $P(A \cap B) = \frac{1}{12}$ and $P(A) \times P(B) = \frac{1}{12}$, the condition for independence holds.

Final Solution: The events A and B are independent events.


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