1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Probability question regarding independent events, where am I wrong ?

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    E1 and E2 are two independent events. The probabilities of an error in judgement by a person regarding E1 and E2 are 0.02 and 0.05 respectively. Find the probability that the person will take the correct decision regarding :
    only one event

    2. Relevant equations
    Two events are independent if occurrence of one is not affected by the other and vice versa .
    C1 be the probability the person judging first one right = (1-0.02) = 0.98
    C2 be the probability the person judging the second one right = (1-0.05) = 0.95

    3. The attempt at a solution

    So if the person judges one event correctly this means the he either judges the first one correctly or the second one correctly
    P(only one event) = P(C1 or C2) = P(C1 + C2) =P(C1) + P(C2) - P(C1).P(C2)
    Which is the wrong anwer . But what's wrong with the logic ?

    The hint shows
    P(only one event) =
    P(E1 is judged correctly and E2 is not judged correctly or E2 is judged correctly and E1 is not judged correctly)

    How are these two logics different I mean the first one also finds the probability of one event judgement to be true . Doesn't it ?

    Correct answer : 0.068
  2. jcsd
  3. Jan 23, 2013 #2
    This equation correctly computes the probability of C1 inclusive or C2, not the probability of C1 exclusive or C2, as the problem asks.

    Think about it like this: C1 includes the "just C1" event as well as the "C1 and C2" event, and C2 includes the "just C2" event as well as the "C1 and C2" event. When you add their probabilities, you double-count the "C1 and C2" event. Because C1 and C2 are independent, the probability of "C1 and C2" is indeed given by P(C1)·P(C2), but subtracting it once from the sum means you're still single-counting the probability of "C1 and C2." You don't want to count it at all, so you need to subtract that term again. You could modify your formula to get the right answer:

    P(C1 xor C2) = P(C1) + P(C2) − 2P(C1)·P(C2)
  4. Jan 23, 2013 #3


    User Avatar
    2017 Award

    Staff: Mentor

    As alternative solution, you can directly evaluate
    P(C1)(1-P(C2)) + (1-P(C1))P(C2)
    As those options are mutually exclusive, you can directly add their probabilites.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook