A coin is tossed thrice and all eight outcomes are assumed equally likely. In which of the following cases are the events A and B independent ?
(i) A : "the first throw results in head"
B : "the last throw results in tail"
(ii) A : "the number of heads is two"
B : "the last throw results in head".

Answer

To determine whether the events A and B are independent in the given scenarios, we will use the definition of independent events. Two events A and B are independent if:

\[ P(A \cap B) = P(A) \times P(B) \]

Let's analyze both cases step by step.

### Case (i):
- **Event A**: "The first throw results in head."
- **Event B**: "The last throw results in tail."

#### Step 1: Determine the total outcomes
When a coin is tossed thrice, the total number of outcomes is \(2^3 = 8\). The outcomes are:
1. HHH
2. HHT
3. HTH
4. HTT
5. THH
6. THT
7. TTH
8. TTT

#### Step 2: Calculate \(P(A)\)
Event A occurs if the first throw is a head. The favorable outcomes are:
- HHH
- HHT
- HTH
- HTT

So, there are 4 favorable outcomes for event A.

\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2} \]

#### Step 3: Calculate \(P(B)\)
Event B occurs if the last throw is a tail. The favorable outcomes are:
- HHT
- HTT
- THT
- TTT

So, there are also 4 favorable outcomes for event B.

\[ P(B) = \frac{\text{Number of favorable outcomes for B}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2} \]

#### Step 4: Calculate \(P(A \cap B)\)
Event \(A \cap B\) occurs if the first throw is a head and the last throw is a tail. The favorable outcomes are:
- HHT
- HTT

So, there are 2 favorable outcomes for \(A \cap B\).

\[ P(A \cap B) = \frac{\text{Number of favorable outcomes for } A \cap B}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4} \]

#### Step 5: Check independence
Now we check if \(P(A \cap B) = P(A) \times P(B)\):

\[ P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent in this case.

### Case (ii):
- **Event A**: "The number of heads is two."
- **Event B**: "The last throw results in head."

#### Step 1: Calculate \(P(A)\)
The outcomes with exactly 2 heads are:
- HHT
- HTH
- THH

So, there are 3 favorable outcomes for event A.

\[ P(A) = \frac{3}{8} \]

#### Step 2: Calculate \(P(B)\)
Event B occurs if the last throw is a head. The favorable outcomes are:
- HHH
- HHT
- HTH

So, there are 3 favorable outcomes for event B.

\[ P(B) = \frac{3}{8} \]

#### Step 3: Calculate \(P(A \cap B)\)
Event \(A \cap B\) occurs if there are exactly 2 heads and the last throw is a head. The favorable outcomes are:
- HHT
- HTH

So, there are 2 favorable outcomes for \(A \cap B\).

\[ P(A \cap B) = \frac{2}{8} = \frac{1}{4} \]

#### Step 4: Check independence
Now we check if \(P(A \cap B) = P(A) \times P(B)\):

\[ P(A) \times P(B) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64} \]

Since \(P(A \cap B) \neq P(A) \times P(B)\), events A and B are not independent in this case.

### Final Conclusion:
- In case (i), events A and B are independent.
- In case (ii), events A and B are not independent.