Part a: Probability $P(WWC)$

The probability of answering a question wrong is denoted as $P(W)$, and the probability of answering a question correctly is denoted as $P(C)$.

• Since there are 5 possible answers, and only one is correct, the probability of guessing correctly is: $P(C) = \frac{1}{5}$

• The probability of guessing wrong is: $P(W) = 1 - P(C) = \frac{4}{5}$

Now, you are asked to find $P(WWC)$, which represents the probability of getting two wrong answers followed by one correct answer. According to the multiplication rule for independent events, we multiply the probabilities of each event occurring:

$P(WWC) = P(W) \times P(W) \times P(C)$

Substituting the probabilities:

$P(WWC) = \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5}$

Now, calculate:

$P(WWC) = \frac{4 \times 4 \times 1}{5 \times 5 \times 5} = \frac{16}{125}$

Thus, $P(WWC) = \frac{16}{125}$

Part b: Probability of getting exactly one correct answer in three guesses

To calculate the probability of getting exactly one correct answer in three guesses, we need to consider all possible ways to get one correct answer and two wrong answers. There are three possible sequences:

1. $CWW$
2. $WCW$
3. $WWC$

Each sequence has the same probability, which we've already calculated for one case ($WWC$) in part a. So, the total probability is:

$P(\text{Exactly one correct}) = P(CWW) + P(WCW) + P(WWC)$

Since each sequence has the same probability of $\frac{16}{125}$, we can multiply it by 3:

$P(\text{Exactly one correct}) = 3 \times \frac{16}{125} = \frac{48}{125}$

Thus, the probability of getting exactly one correct answer when making three guesses is:

$P(\text{Exactly one correct}) = \frac{48}{125}$

Summary:

• $P(WWC) = \frac{16}{125}$
• $P(\text{Exactly one correct}) = \frac{48}{125}$

Would you like more details on any part of the solution?

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Here are 5 related questions you can explore:

1. What is the probability of getting no correct answers in three guesses?
2. What is the probability of getting exactly two correct answers in three guesses?
3. How does the probability change if there are four possible answers instead of five?
4. What is the probability of getting all three answers correct?
5. How would the probabilities change if you could eliminate one incorrect answer before guessing?

Tip: When dealing with probabilities of independent events, remember to use the multiplication rule for each sequence of outcomes.