Question

What is the probability of randomly selecting a student who earned an A, given that the student studied?

Answer 1

I think its 6/7 but I have no idea if it's independent or dependent

Answer 2

You can test if two events are independent by checking if \[P(A \text{ and } B) = P(A)\cdot P(B)\]

Answer 3

I'm still trying to make sense of the table, the column headers are a bit confusing...

Answer 4

I'm sure it goes like this

Answer 5

OK, that makes more sense.

Answer 6

I don't really understand what A and B stand for

Answer 7

A 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

Answer 8

So, 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

Answer 9

So, 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.

Answer 10

aaah okay okay thanks i was trying to figure that out

Answer 11

Cool, glad it helped! So, how about the probability of selecting a student who studied? P(B)?

Answer 12

1/34?

Answer 13

wait nvm hold on

Answer 14

It's a little tricky, let's start with this first, how many students studied total? only 1? I think it's more. :)

Answer 15

24/34?

Answer 16

Getting 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.

Answer 17

27/34?

Answer 18

Close. 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. :)

Answer 19

Here's how I look at it.

Answer 20

7/34??
i know it seems like i'm guessing but i'm really not

Answer 21

No, no, I know you are trying to readjust your interpretations based on my responses. Here's how I look at it.

Answer 22

WAIT 30/34?

Answer 23

AHA! There you go!

Answer 24

The bottom total row totals the data for the columns. The right total column totals the data for the rows. It's annoying.

Answer 25

Now, for P(A and B) we want the number of students who studied (yes column) AND who got an A (yes row).

Answer 26

7/34? for studied and 6/34 who got an A???

Answer 27

Just one probability. The number of students who both studied AND got an A is 6. Thus,
P(A and B) = 6/34.

Answer 28

aaaah okay

Answer 29

So far we have P(A) = 7/34, P(B) = 30/34, and P(A and B) = 6/34.

Answer 30

yeah so 6/34=7/34*30/34

Answer 31

Will be true ONLY if they really are independent, otherwise this will be false.

Answer 32

You 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. :)

Answer 33

Are you suggesting it's 1/5?? I don't see how you could get that as an answer

Answer 34

Yep, that's what I'm suggesting. It's not obvious. But here's how I think about it.

Answer 35

We 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...)

Answer 36

Of these 30 students, only 6 earned A's, thus the probability of A given B is

P(A|B) = 6/30

Answer 37

What you calculated is the probability that the student studied GIVEN that they earned an A. You did the reverse probability.

Answer 38

aaaah okay

Answer 39

Hopefully 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. :)