Mathematics, 27.02.2020 19:22 thealienminecra

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.

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Answer from: margaret1758

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.

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

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

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.

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

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

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

Answer from: Quest

2.83941151 would be the answer you're looking for! : )

Answer from: Quest

Idon’t know what the answer is i wish i could

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