A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

briana.img

  • one year ago

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

  • This Question is Open
  1. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm sure it goes like this

  5. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    OK, that makes more sense.

  6. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  8. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  9. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  10. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aaah okay okay thanks i was trying to figure that out

  11. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  12. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1/34?

  13. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait nvm hold on

  14. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  15. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    24/34?

  16. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  17. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    27/34?

  18. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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. :)

  19. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here's how I look at it.

  20. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  22. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    WAIT 30/34?

  23. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    AHA! There you go!

  24. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  25. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  26. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  27. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  28. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aaaah okay

  29. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  30. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  31. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  32. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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. :)

  33. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  34. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  35. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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...)

  36. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Of these 30 students, only 6 earned A's, thus the probability of A given B is P(A|B) = 6/30

  37. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  38. briana.img
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aaaah okay

  39. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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. :)

  40. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.