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: Using probability union

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data

    If 60% of households subscribe to Metro(M) newspaper, 80% subscribe to local (L) newspaper, and 50% subscribe to both,

    1)what's the probability that a random household subscribes to at least one paper?
    2) what's the probability that a random household subscribes to exactly one paper?

    3. The attempt at a solution

    1) The probability of at least one paper subscribed is P(M U L) = P(M) + P(L) - P(M AND L) which gives the answer of 0.9. But I'm wondering why this works? The phrasing "at least" means the possibilities can be just P(M), just P(L), or P(M AND L) but from the equation we are substracting out P(M AND L), which means P(M) and P(L) are the only possible outcomes. What's wrong with my thinking here?

    2) Probability = P(Mc and L) U P(Mc and L) and they are mutually exclusive so you just sum the two probabilities.

    Pc = 1- P
    Not sure how to find P(Mc and L) since they aren't independent.

  2. jcsd
  3. Jun 12, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    P(L & M) is part of P(M) and is also part of P(L), so the summation P(L) + P(M) counts P(L&M) twice. That is why you need to subtract it---so you only count it once.

  4. Jun 13, 2012 #3


    User Avatar
    Science Advisor

    Imagine that there are 100 households. Then 60 subscribe to M and 50 subscribe to both so 10 subscribe to M only. 80 subscribe to L and 50 subscribe to both so 30 subscribe to L only. That makes a total of 10+ 50+ 30= 90 households out of 100 that subscribe to at least one newspaper and 10+ 30= 40 that subscribe to exactly one newspaper.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook