Let's break down the problem and use the provided information to solve the questions step by step. We will use the law of total probability and Bayes' theorem to answer the questions.

Given Information:

• Proportion of population in each state:

  • $P(X) = 0.25$ (State X)
  • $P(Y) = 0.45$ (State Y)
  • $P(Z) = 0.30$ (State Z)

• Probability of infection in each state:

  • $P(C|X) = 0.20$ (State X)
  • $P(C|Y) = 0.10$ (State Y)
  • $P(C|Z) = 0.15$ (State Z)

Let $C$ represent the event that a citizen has been infected with Covid-19, and let $C^c$ represent the event that a citizen has not been infected with Covid-19.

(a) Probability that a randomly selected citizen has not been infected with Covid-19 and lives in state X

To find this probability, we need to find the joint probability $P(C^c \cap X)$. This can be found using:

$P(C^c \cap X) = P(X) \times P(C^c | X)$

Since $P(C^c | X) = 1 - P(C | X) = 1 - 0.20 = 0.80$, we get:

$P(C^c \cap X) = 0.25 \times 0.80 = 0.20$

So, the probability that a randomly selected citizen has not been infected with Covid-19 and lives in state X is 0.20.

(b) Probability that a randomly selected citizen has been infected with Covid-19

To find the total probability that a randomly selected citizen has been infected with Covid-19, we use the law of total probability:

$P(C) = P(C | X) \times P(X) + P(C | Y) \times P(Y) + P(C | Z) \times P(Z)$

Substituting the given values:

$P(C) = (0.20 \times 0.25) + (0.10 \times 0.45) + (0.15 \times 0.30)$

Let's calculate it step by step:

$P(C) = 0.05 + 0.045 + 0.045 = 0.14$

So, the probability that a randomly selected citizen has been infected with Covid-19 is 0.14.

(c) Given that a randomly selected citizen has been infected with Covid-19, find the probability that the selected citizen is from state Y

This is a conditional probability question. We need to find $P(Y | C)$, which can be found using Bayes' theorem:

$P(Y | C) = \frac{P(C \cap Y)}{P(C)}$

Where $P(C \cap Y) = P(C | Y) \times P(Y)$, so:

$P(Y | C) = \frac{P(C | Y) \times P(Y)}{P(C)} = \frac{0.10 \times 0.45}{0.14}$

Let's calculate it:

$P(Y | C) = \frac{0.045}{0.14} \approx 0.321$

So, the probability that a randomly selected citizen is from state Y, given that they have been infected with Covid-19, is approximately 0.321 or 32.1%.

Final Answers:

• (a) The probability that a randomly selected citizen has not been infected with Covid-19 and lives in state X is 0.20.
• (b) The probability that a randomly selected citizen has been infected with Covid-19 is 0.14.
• (c) The probability that a randomly selected citizen is from state Y, given that they have been infected with Covid-19, is approximately 0.321.

Would you like further explanations or details on any of these steps?

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Here are 5 related questions you can explore:

1. What is the probability that a randomly selected citizen has not been infected with Covid-19 and lives in state Z?
2. Given that a randomly selected citizen has not been infected with Covid-19, what is the probability that they are from state X?
3. What is the probability that a randomly selected citizen is from state Z, given that they have been infected with Covid-19?
4. Find the probability that a randomly selected citizen has been infected with Covid-19 and lives in state Z.
5. What is the probability that a randomly selected citizen is from state Y, given that they have not been infected with Covid-19?

Tip: When dealing with conditional probabilities, Bayes' theorem is a powerful tool to find the reverse conditional probabilities.