- Thread starter
- #1

- Thread starter f666
- Start date

- Thread starter
- #1

- Jan 30, 2018

- 368

Imagine 1000 people in the coffee shop. 20% of them, 200, are "students coming for Dan's office hours" and the other 80%, 800, are not.

Of the 200 'students coming for Dan's office hours", 20%, 0.2(200)= 40 students, got hot chocolate, 50%, 0.5(200)= 100 students, got nothing, and I presume that the remaining 60 students got coffee.

Of the 800 people what are NOT "students coming from Dan's office hours", 65%, 0.65(800)= 520 people, got coffee, 10%, 0.10(800)= 80 people, got hot chocolate, and 25%, 0.25(800)= 200 people, got nothing. (That adds to 800 people so our assumption that "coffee", "hot chocolate", or "nothing" are the only options is valid.)

"What are the chances that someone in the shop during those two hours was a student who came in for office hours given that they got hot chocolate?"

From above, a total of 40+ 80= 120 people got hot chocolate. 40 of those were "a student who came in for officice hours" so the probability that "someone in the shop during those two hours was a student who came in for office hours given that they got hot chocolate" is 40/120= 1/3.

- Thread starter
- #3