An instructor surveyed her class of 45 students and found that most of them watc

Question

An instructor surveyed her class of 45 students and found that most of them watched TV and/or got on the Internet last night. 10 students said they watched TV but did not get on the Internet. 16 students said they got on the Web but did not watch TV last night. 43 of the students watched TV or got on the Internet (or both) last night. What is the probability that a single randomly-selected student got on the Internet last night? Show work. Let T ={students who watched TV) and W ={students who got on the Internet}. Determine the number of attendees belonging to each of the regions I, II, Ill, IV.

Explanation / Answer

a) Out of 45 students 43 students either watched TV or got on the internet.

Oly (45-43) = 2 students did nothing.

10 students watched TV but did not get on the internet.

So in total (10+2) = 12 stdents did not get on the internet.

So (45-12) =33 students got on the internet last night.

So, the probability that a single randomly selected students got on the internet = 33 / 45 = 0.73. (Ans).

b) Number of students either watch TV or got on internet =43.

Number of students who did nothing = 45-2 43.

Number of students only watched TV = 10.

Number of students only got on internet = 16.

Number of students who watched TV = (43-16) = 27.

Number of students who got on internet = (43-10) = 33.

Number of students who did both = (43 - 10 - 16) = 17.

Hence Region I = Number of students who only watched TV =10.

Region II = Number of students who did both = 17.

Region III = Number of students who only got on internet = 16.

Region IV = Number of students who did nothing = 2..

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