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briana.img
 one year ago
What is the probability of randomly selecting a student who earned an A, given that the student studied?
briana.img
 one year ago
What is the probability of randomly selecting a student who earned an A, given that the student studied?

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briana.img
 one year ago
Best ResponseYou've already chosen the best response.0I think its 6/7 but I have no idea if it's independent or dependent

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1You can test if two events are independent by checking if \[P(A \text{ and } B) = P(A)\cdot P(B)\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I'm still trying to make sense of the table, the column headers are a bit confusing...

briana.img
 one year ago
Best ResponseYou've already chosen the best response.0I'm sure it goes like this

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, that makes more sense.

briana.img
 one year ago
Best ResponseYou've already chosen the best response.0I don't really understand what A and B stand for

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1A and B are two different events. You can define them however you want, but I would suggest. A: Student gets an A B: Student studied

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1So, the independence test says our two events are independent if and only if the probability of A and B = probability of A * probability of B

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1So, we first figure out what is the probability of selecting a student who got an A? (this is P(A) ). 7 students earned A's out of 34 students total so P(A) = 7/34.

briana.img
 one year ago
Best ResponseYou've already chosen the best response.0aaah okay okay thanks i was trying to figure that out

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Cool, glad it helped! So, how about the probability of selecting a student who studied? P(B)?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1It's a little tricky, let's start with this first, how many students studied total? only 1? I think it's more. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Getting warmer. That's the number of students who studied (the yes column) AND did not get an A (the no row). We want ALL the students who stuided (whether they got an A or not). Just cover up the "yes/no" rows on the left and focus on the totals.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Close. I know... now I'm just torturing you... :) That's the total number of students who did not get an A. Interpreting the table is half of the battle here. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Here's how I look at it.

briana.img
 one year ago
Best ResponseYou've already chosen the best response.07/34?? i know it seems like i'm guessing but i'm really not

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1No, no, I know you are trying to readjust your interpretations based on my responses. Here's how I look at it.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1The bottom total row totals the data for the columns. The right total column totals the data for the rows. It's annoying.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Now, for P(A and B) we want the number of students who studied (yes column) AND who got an A (yes row).

briana.img
 one year ago
Best ResponseYou've already chosen the best response.07/34? for studied and 6/34 who got an A???

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Just one probability. The number of students who both studied AND got an A is 6. Thus, P(A and B) = 6/34.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1So far we have P(A) = 7/34, P(B) = 30/34, and P(A and B) = 6/34.

briana.img
 one year ago
Best ResponseYou've already chosen the best response.0yeah so 6/34=7/34*30/34

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Will be true ONLY if they really are independent, otherwise this will be false.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1You should calculate and see that these are NOT EQUAL. So, the events are NOT independent. They are DEPENDENT. You may want to also check your calculation on students getting an A, given they studied. :)

briana.img
 one year ago
Best ResponseYou've already chosen the best response.0Are you suggesting it's 1/5?? I don't see how you could get that as an answer

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yep, that's what I'm suggesting. It's not obvious. But here's how I think about it.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1We are GIVEN that the students studied. There are only 30 students who did this according to the table. Our total "pool" of students has shrunk from 34, down to 30. We only care about these students (that sounds kind of mean...)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Of these 30 students, only 6 earned A's, thus the probability of A given B is P(AB) = 6/30

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1What you calculated is the probability that the student studied GIVEN that they earned an A. You did the reverse probability.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Hopefully that helps you to make sense of this table. It's not super easy, but it isn't super hard either. The trick is just interpreting the table and the data correctly. Of course, if you were making your own table, you would have written the data down in a way that makes sense to you rather than using someone else's organization. :)
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