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Statistics Problem

  1. Jun 26, 2009 #1
    The exercisie is in the attachment.

    I am trying to solve it using a Binomial distribution, using n = 64, and p = 0.2.

    P(X=1) =

    (64) (0.2)^1 (0.8)^63
    ( 1)

    and, I get that P(X=1) = 1.00434 * 10^-5

    However, this does not seems correct, can anyone help me here?? I need to submit this in 4hours, so please help me. Thanks.

    Attached Files:

  2. jcsd
  3. Jun 26, 2009 #2
    Could the answer be: 0.030260441?
  4. Jun 26, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How did you calculate that number?
  5. Jun 26, 2009 #4


    Staff: Mentor

    Seems OK to me. As you described it, you're looking for the probability of 1 success in 64 trials, where each trial has a 0.2 probability of success.

    Another way to think about this is that you have a jar with 5 balls in it, with one white ball and four black balls. For each trial, you reach in (without looking) and pick out a ball and record its color, then put the ball back in. If you do this 64 times, what's the probability that you get 1 white ball out of 64 trials? It's pretty low.
  6. Jun 26, 2009 #5


    Staff: Mentor

    Now that I can see your attachment, what you calculated seems correct for P(X = 1), but this isn't the probability that the problem asks for, which is the probability that at least one passenger will get bumped.
  7. Jun 26, 2009 #6
    Thanks for the input so far guys:

    As for my last attempt:

    After I careful read the problem, it says "at least one passenger..", so I assumed that it was either 1 passager needs to be "bumped", or 2, or 3, or 4, or 5, or 6.

    So, I added the probabilities of:

    P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

    And that's how I got the 0.030260441.
  8. Jun 26, 2009 #7
    Yes, Mark44! Is it correct now??
  9. Jun 26, 2009 #8


    Staff: Mentor

    Looks OK, assuming you calculated the individual probabilities correctly. For example, for P(X = 2) you have 64C2*(.2)2*(.8)62, right? (That number at the beginning works out to 64*63/2.)
  10. Jun 26, 2009 #9
    Thank you, Mark. I did it with a formula using Excel, so I'm pretty sure it's correct. Thanks a lot!
  11. Jun 26, 2009 #10
    Actually, I had an error there, the correct answer is: 0.027605266.

    My bad.
  12. Jun 26, 2009 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Anyways, now that you've worked it out, it might interest you to know there's a cute shortcut -- what is the probability that nobody gets bumped?
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