Question

A survey showed that 24% of college students read newspapers on a regular basis and that 82% 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.

(a) 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.
(b) 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.

Answer 1

Answer:

a) 0.88 = 88% probability that a student watches TV news regularly, given that he or she regularly reads newspapers.

b) 0.26 = 26% probability that a randomly selected college student reads newspapers regularly, given that he or she watches TV news regularly

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a college student read newspapers on a regular basis.

B is the probability that a college student regularly watch the news of TV.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a students reads newspapers but does not watches TV and $A \cap B$ is the probability that a student does both of these things.

By the same logic, we have that:

$B = b + (A \cap B)$

The survey also showed that 21% of college students both follow TV news regularly and read newspapers regularly.

This means that $A \cap B = 0.21$

24% of college students read newspapers on a regular basis;

This means that $A = 0.24$.

82% of college students regularly watch the news on TV.

This means that $B = 0.82$

(a) 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.

By the Bayes Rule, probability of event B, given that A, is given by the following formula.

$P(B|A) = \frac{A \cap B}{A}$

So

$P(B|A) = \frac{0.21}{0.24} = 0.88$

0.88 = 88% probability that a student watches TV news regularly, given that he or she regularly reads newspapers.

(b) 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.

By the Bayes Rule, probability of event A, given that B, is given by the following formula.

$P(A|B) = \frac{A \cap B}{B}$

So

$P(A|B) = \frac{0.21}{0.82} = 0.26$

0.26 = 26% probability that a randomly selected college student reads newspapers regularly, given that he or she watches TV news regularly