The image shows a probability problem involving a coin being tossed three times, with each outcome equally likely. The problem involves analyzing various events based on the outcomes of the tosses, such as getting two heads or ensuring not all tosses are the same. Let's break down the solution step by step:

Part (a): Determine the outcomes and probability of each event

• The possible outcomes of tossing a coin three times are:
  • $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$ (Total: 8 outcomes)
  • Each outcome has a probability of $\frac{1}{8}$.

For the events described:

• Event A: Exactly two tosses are heads.

  • The outcomes with exactly two heads are: $HHT, HTH, THH$.
  • Probability: $\frac{3}{8}$.

• Event B: Not all tosses are the same (i.e., the outcome is not $HHH$ or $TTT$).

  • The outcomes where not all tosses are the same are: $HHT, HTH, HTT, THH, THT, TTH$.
  • Probability: $\frac{6}{8} = \frac{3}{4}$.

• Event A and B: Exactly two heads and not all tosses are the same.

  • The outcomes for Event A are $HHT, HTH, THH$, and these outcomes are also in Event B (since none of them are $HHH$ or $TTT$).
  • Probability: $\frac{3}{8}$.

Part (b): Given Event B, what is the probability of Event A?

Here, we use conditional probability:

• $P(A | B) = \frac{P(A \text{ and } B)}{P(B)}$
• $P(A \text{ and } B) = \frac{3}{8}$
• $P(B) = \frac{6}{8} = \frac{3}{4}$

Thus: $P(A | B) = \frac{\frac{3}{8}}{\frac{3}{4}} = \frac{3}{8} \times \frac{4}{3} = \frac{1}{2}$

Part (c): Calculate the following probabilities

• $\frac{P(A \text{ and } B)}{P(B)}$:

  • As found in Part (b), this is $\frac{1}{2}$.

• $P(A | B)$:

  • Already calculated, this is $\frac{1}{2}$.

If you need more details or have any questions about this, feel free to ask!

Related questions:

1. How is the probability of a single outcome in coin tossing calculated?
2. What is the difference between independent and mutually exclusive events in probability?
3. How does conditional probability differ from regular probability?
4. Can you provide an example where $P(A | B)$ is different from $P(B | A)$?
5. What is Bayes' theorem, and how does it relate to conditional probability?

Tip:

When dealing with conditional probability, always check if the total probability space changes, as you are focusing on a subset of all possible outcomes.