1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Mark44

    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

    Mark44

    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

    Mark44

    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

    Hurkyl

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Statistics Problem
  1. Statistic problem (Replies: 2)

  2. Statistics Problem (Replies: 6)

  3. Statistic Problem (Replies: 0)

  4. Statistics problem (Replies: 1)

Loading...