+ Reply to Thread
Results 1 to 7 of 7

Thread: Probability Help

  1. #1
    Points: 29, Level: 1
    Level completed: 58%, Points required for next Level: 21

    Posts
    3
    Thanks
    2
    Thanked 0 Times in 0 Posts

    Probability Help




    Hi,

    I giving my IGCSE this November and came across a problem that I can't solve. It'd be really helpful if someone can help me with the sum.

    " A square board measuring 30 cm by 30 cm has two rectangular cards attached to it. The large card measures 15 cm by 20 cm and the small card measures 8 cm by 12 cm. The large card covers exactly one-quarter of the small card. If a dart is thrown at the board, what is the probability of piercing:

    a) both cards
    b) both cards, given that it pierces at least one of them"

    I got the answer to question a) by finding the intersection

    1/4 x [(8x12) / (30x30)] = 24 / 900 = 2/75


    but cannot find anyway to solve question b).
    Attached Images  
    Last edited by preziano; 06-09-2014 at 02:31 AM.

  2. #2
    Points: 4,841, Level: 44
    Level completed: 46%, Points required for next Level: 109
    Con-Tester's Avatar
    Posts
    167
    Thanks
    3
    Thanked 59 Times in 57 Posts

    Re: Probability Help

    You’ll need to use the conditional probability formula. In the formula, outcome A = “dart pierces both cards”, and outcome B = “dart pierces at least one card”. The probabilities are directly related to the surface areas of the four possible regions (no card, card 1, card2, both cards) the dart can stick in.

    (It is worth noting that the whole problem is already one of conditional probability because it implicitly assumes that the dart sticks somewhere in the 30 × 30 cm board.)

  3. The Following User Says Thank You to Con-Tester For This Useful Post:

    preziano (06-09-2014)

  4. #3
    Points: 29, Level: 1
    Level completed: 58%, Points required for next Level: 21

    Posts
    3
    Thanks
    2
    Thanked 0 Times in 0 Posts

    Re: Probability Help

    Thanks for the response, I cant seem to understand what the question truly means. What is the probability that the darts hits both the cards given that it pierces at least one of them???

  5. #4
    Points: 4,841, Level: 44
    Level completed: 46%, Points required for next Level: 109
    Con-Tester's Avatar
    Posts
    167
    Thanks
    3
    Thanked 59 Times in 57 Posts

    Re: Probability Help

    Refer to the Wikipedia article I linked to (click on the blue “conditional probability” text). Do you understand conditional probability, specifically what the formula “P(A|B) = P(A ∩ B)/P(B)” means?

    As I suggested earlier, use outcome A = “dart pierces both cards”, and outcome B = “dart pierces at least one card”. From your diagram, the following should be clear:

    (1) P(A ∩ B) is the probability that the dart pierces both cards and it pierces at least one card. But if it pierces both, then it is also true that it pierces at least one of them. Therefore, outcome A subsumes outcome B, so that P(A ∩ B) = P(A), which you have already calculated as 24/900.

    (2) P(B) is true if the dart hits any card or both of them. The total area covered by the cards is 20×15+(3/4)×12×8 cm² = 372 cm² out of 900 cm² of the board. Therefore, P(B) = 372/900.

    Putting these values in the formula gives P(A|B) = P(A ∩ B)/P(B) = (24/900)/(372/900) = 24/372 = 2/31.

    ____________________________________________

    Edit: What the question means is the following: Take all of the dart throws at the board. From them, select only those throws where the dart pierced at least one of the cards. In that subset, what proportion pierced both cards?
    Last edited by Con-Tester; 06-09-2014 at 03:24 AM. Reason: Added explanation.

  6. The Following User Says Thank You to Con-Tester For This Useful Post:

    preziano (06-09-2014)

  7. #5
    Points: 3,666, Level: 38
    Level completed: 11%, Points required for next Level: 134

    Posts
    48
    Thanks
    3
    Thanked 5 Times in 5 Posts

    Re: Probability Help

    Quote Originally Posted by preziano View Post
    Thanks for the response, I cant seem to understand what the question truly means. What is the probability that the darts hits both the cards given that it pierces at least one of them???
    Maybe this example will help?

    My Aunt is going to come and visit me at some point next week, with an equal likelihood of any day.

    Q. What is the probability that she will visit on a Wednesday?
    A. 1/7

    Q. What is the probability that she will visit on a Saturday?
    A. 1/7

    New information: My aunt has now told me she will be visiting on a weekday.

    Q. What is the probability that she will visit on a Wednesday, given she will be visiting on a weekday?
    A. 1/5

    Q. What is the probability that she will visit on a Saturday, given she will be visiting on a weekday?
    A. 0

    The "given" in some way limits or clarifies the situation which the question is asking about.

  8. #6
    Points: 29, Level: 1
    Level completed: 58%, Points required for next Level: 21

    Posts
    3
    Thanks
    2
    Thanked 0 Times in 0 Posts

    Re: Probability Help

    That was really helpful. I kept dividing (2/75) [the answer of both cards] with (372/900) [any card or both]. Thanks a lot for clearing my doubt.

  9. #7
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Probability Help


    Yes. As mentioned above, the information is "shrinking" the sample space from the 30 x 30 board to the union of two cards.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats