Question

conditional probability: basic

a survey showed that 23% of college students read newspapers on a regular basis and that 83% of college students regularly watch the news on tv. the survey also showed that 21% of college students both follow tv news regularly and read newspapers regularly.

answer the questions below. (if necessary, consult a list of formulas.)

(a) what is the probability that a randomly selected college student reads newspapers regularly, given that he or she watches tv news regularly? round your answer to 2 decimal places.

(b) what is the probability that a student watches tv news regularly, given that he or she regularly reads newspapers? round your answer to 2 decimal places.

Explanation:

Define the given probabilities

We define the events based on the survey data:

• Let \(N\) be the event that a student reads newspapers regularly.
• Let \(T\) be the event that a student watches TV news regularly.

From the problem statement, we have:

• \(P(N) = 23\% = 0.23\)
• \(P(T) = 83\% = 0.83\)
• \(P(N \cap T) = 21\% = 0.21\)

Calculate conditional probability for part (a)

Using the Conditional Probability formula:

$$ P(N \mid T) = \frac{P(N \cap T)}{P(T)} $$

Substitute the values:

$$ P(N \mid T) = \frac{0.21}{0.83} \approx 0.2530 $$

Rounding to 2 decimal places gives \(0.25\).

Calculate conditional probability for part (b)

Using the Conditional Probability formula:

$$ P(T \mid N) = \frac{P(T \cap N)}{P(N)} $$

Substitute the values:

$$ P(T \mid N) = \frac{0.21}{0.23} \approx 0.9130 $$

Rounding to 2 decimal places gives \(0.91\).

Answer:

Question 1

What is the probability that a randomly selected college student reads newspapers regularly, given that he or she watches TV news regularly? Round your answer to 2 decimal places.
<blank>0.25</blank>

Question 2

What is the probability that a student watches TV news regularly, given that he or she regularly reads newspapers? Round your answer to 2 decimal places.
<blank>0.91</blank>