Problem 4) A fair coin is tossed twice. Let E, be the event that both tosses have the same outcome, that is, E1 = (HH, TT). Let E2 be the event that the first toss is a head, that is, E2 = (HH, HT). Let Ez be the event that the second toss is a head, that is, E3 = (TH, HH). Show that E1, E2, and E3 are pairwise independent but not mutually independent. %3D %3D

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**Problem 4) A fair coin is tossed twice.**

Let \(E_1\) be the event that both tosses have the same outcome, that is, \(E_1 = (HH, TT)\).

Let \(E_2\) be the event that the first toss is a head, that is, \(E_2 = (HH, HT)\).

Let \(E_3\) be the event that the second toss is a head, that is, \(E_3 = (TH, HH)\).

Show that \(E_1\), \(E_2\), and \(E_3\) are pairwise independent but not mutually independent.

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### Explanation for Educational Purposes:

When a fair coin is tossed twice, there are four possible outcomes: \(HH\) (both heads), \(HT\) (head then tail), \(TH\) (tail then head), and \(TT\) (both tails).

### Event Definitions:

1. **Event \(E_1\)**: Both tosses have the same outcome.
   - Possible outcomes: \(HH\), \(TT\)
   - Hence, \(E_1 = \{ HH, TT \}\)
   
2. **Event \(E_2\)**: The first toss is a head.
   - Possible outcomes: \(HH\), \(HT\)
   - Hence, \(E_2 = \{ HH, HT \}\)
   
3. **Event \(E_3\)**: The second toss is a head.
   - Possible outcomes: \(TH\), \(HH\)
   - Hence, \(E_3 = \{ TH, HH \}\)

### Independence Check:

- **Pairwise Independence**: This means that any two of these events are independent of each other.
- **Mutual Independence**: This means that all three events are independent, which includes the condition that the probability of the intersection of all three events equals the product of their individual probabilities.

To show pairwise independence:

1. \(E_1\) and \(E_2\) are independent.
2. \(E_1\) and \(E_3\) are independent.
3. \(E_2\) and \(E_3\) are independent.

### Conclusion:

While \(E_1\), \(E_2\), and \(E_3\) are pairwise independent (
Transcribed Image Text:--- **Problem 4) A fair coin is tossed twice.** Let \(E_1\) be the event that both tosses have the same outcome, that is, \(E_1 = (HH, TT)\). Let \(E_2\) be the event that the first toss is a head, that is, \(E_2 = (HH, HT)\). Let \(E_3\) be the event that the second toss is a head, that is, \(E_3 = (TH, HH)\). Show that \(E_1\), \(E_2\), and \(E_3\) are pairwise independent but not mutually independent. --- ### Explanation for Educational Purposes: When a fair coin is tossed twice, there are four possible outcomes: \(HH\) (both heads), \(HT\) (head then tail), \(TH\) (tail then head), and \(TT\) (both tails). ### Event Definitions: 1. **Event \(E_1\)**: Both tosses have the same outcome. - Possible outcomes: \(HH\), \(TT\) - Hence, \(E_1 = \{ HH, TT \}\) 2. **Event \(E_2\)**: The first toss is a head. - Possible outcomes: \(HH\), \(HT\) - Hence, \(E_2 = \{ HH, HT \}\) 3. **Event \(E_3\)**: The second toss is a head. - Possible outcomes: \(TH\), \(HH\) - Hence, \(E_3 = \{ TH, HH \}\) ### Independence Check: - **Pairwise Independence**: This means that any two of these events are independent of each other. - **Mutual Independence**: This means that all three events are independent, which includes the condition that the probability of the intersection of all three events equals the product of their individual probabilities. To show pairwise independence: 1. \(E_1\) and \(E_2\) are independent. 2. \(E_1\) and \(E_3\) are independent. 3. \(E_2\) and \(E_3\) are independent. ### Conclusion: While \(E_1\), \(E_2\), and \(E_3\) are pairwise independent (
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