Question

1. In a certain population, 26% of the persons smoke and 9% have a certain type of heart disease.Moreover, 14% of the persons who smoke have the disease.

(a) What percentage of the population smoke and have the disease?

(b) What percentage of the population with the disease also smoke?

2. At a certain university in the United States, 64% of the students are at least bilingual{speaking English and at least one other language. Of these students, 79% speak Spanish and, of the 79% who speak Spanish, 12% also speak French. Determine the probability that a randomly selected student at this university:

(a) does not speak Spanish.

(b) speaks Spanish and French.

Answer 1

Let S represent the event that an individual smokes & D represent the event that an individual has the disease

Answer 1a):

In this case we have to find P(D ∩ S)

P(S) = 0.26

P(D | S) = 0.14

P(D ∩ S) = P(D | S)*P(S) = 0.14*0.26

P(D ∩ S) = 0.0364

Percentage of the population who smoke and have the disease = 0.0364*100 = 3.64%

Answer 1b):

In this case, we have to find P(S | D)

P(S|D) = P(S ∩ D)/P(D)

Given, P(D) = 0.09 and P(S ∩ D) = 0.0364 (Refer Part a)

P(S|D) = 0.0364/0.09

P(S|D) = 0.4044

Percentage of the population with the disease also smoke = 0.4044*100 = 40.44%

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