An honest coin is tossed 8 times in a row. Find the probability of each of the following events.
​(a) Upper E 1E1​: ​"None of the tosses comes up​ tails."
​(b) Upper E 2E2​: ​"Exactly one of the 8 tosses comes up​ tails."
​(c) Upper E 3E3​: ​"7 or more of the tosses come up​ tails.

We are asked to find the probabilities of specific events occurring in a series of 8 coin tosses, assuming the coin is fair (i.e., each toss has a 50% chance of landing heads or tails). 

### Key points:
- Each toss is independent, and there are 2 possible outcomes for each toss (heads or tails).
- The total number of possible outcomes after 8 tosses is \(2^8 = 256\).

#### (a) \( E_1 \): "None of the tosses comes up tails."
This event implies that all 8 tosses come up heads. Since each toss has a \( \frac{1}{2} \) probability of being heads, the probability for 8 consecutive heads is:
\[
P(E_1) = \left( \frac{1}{2} \right)^8 = \frac{1}{256}
\]
Thus, the probability of no tails (all heads) is \( \frac{1}{256} \).

#### (b) \( E_2 \): "Exactly one of the 8 tosses comes up tails."
This event implies that 1 out of the 8 tosses is tails, and the other 7 tosses are heads. The number of ways to choose 1 tail out of 8 tosses is given by the binomial coefficient \( \binom{8}{1} \), and the probability of any specific arrangement with 1 tail and 7 heads is \( \left( \frac{1}{2} \right)^8 \). 

The binomial coefficient \( \binom{8}{1} \) is:
\[
\binom{8}{1} = 8
\]
Thus, the probability of exactly one tail is:
\[
P(E_2) = \binom{8}{1} \left( \frac{1}{2} \right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} = \frac{1}{32}
\]

#### (c) \( E_3 \): "7 or more of the tosses come up tails."
This event includes two cases:
- **Case 1**: Exactly 7 tails (1 head).
- **Case 2**: Exactly 8 tails (0 heads).

For **Case 1**, the number of ways to choose 7 tails out of 8 tosses is given by \( \binom{8}{7} = 8 \). The probability for this case is:
\[
P(\text{7 tails}) = \binom{8}{7} \left( \frac{1}{2} \right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} = \frac{1}{32}
\]

For **Case 2**, there is only 1 way to have 8 tails (i.e., \( \binom{8}{8} = 1 \)). The probability for this case is:
\[
P(\text{8 tails}) = \binom{8}{8} \left( \frac{1}{2} \right)^8 = 1 \times \frac{1}{256} = \frac{1}{256}
\]

The total probability for \( E_3 \) (7 or more tails) is the sum of these two cases:
\[
P(E_3) = P(\text{7 tails}) + P(\text{8 tails}) = \frac{1}{32} + \frac{1}{256}
\]
To add these, we express both fractions with a common denominator:
\[
P(E_3) = \frac{8}{256} + \frac{1}{256} = \frac{9}{256}
\]

### Final Answers:
(a) \( P(E_1) = \frac{1}{256} \)  
(b) \( P(E_2) = \frac{1}{32} \)  
(c) \( P(E_3) = \frac{9}{256} \)

Would you like more details on any of these calculations, or do you have any other questions?

Here are five related questions for further practice:
1. What is the probability of getting exactly 4 heads in 8 tosses?
2. How many different outcomes are there with exactly 3 tails in 8 tosses?
3. If the coin is biased with \( P(\text{heads}) = 0.6 \), how does that affect the probability of getting exactly 5 heads?
4. What is the probability of getting at least 6 heads in 10 tosses?
5. How many sequences of 10 tosses result in exactly 7 heads?

**Tip:** For problems involving multiple independent trials, the binomial distribution is a useful tool to calculate probabilities.

An honest coin is tossed 8 times in a row. Find the probability of each of the following events.
(a) E1: "None of the tosses comes up tails."
(b) E2: "Exactly one of the 8 tosses comes up tails."
(c) E3: "7 or more of the tosses come up tails."

Learn how to calculate the probability of various coin toss events in 8 tosses, including no tails, exactly one tail, and seven or more tails. Using concepts from probability and combinatorics, this problem is ideal for students in Grades 9-12.

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