Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Percentage Question

  1. Dec 27, 2004 #1
    Hey all, I'm a lawyer with some interest in physics, however, I absolutely suck at math so I've come here for help.

    Here's the question. I've got this case whose success depends on the occurence of two separate conditions. This case can not be won if BOTH conditions do not occur.

    Condition 1 has approximately a 60% chance of occurence
    Condition 2 has approximately a 70% chance of occurence

    What is the total probability that this case will be won? How would I go about figuring this out in the future?
     
  2. jcsd
  3. Dec 27, 2004 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    You cannot answer the question unless you know if the conditions are indpendent. you might also wish to remove some of the negatives - we aren't lawyers, unnecessary double negatives are not attractive.

    for what it's worth, the probability of both 1 and 2 not occuring is .4 times .3 if they are independent. or the probability of either one happening is 0.6+0.7 - P(both occur). but if we do not know about the independence or not we're stuffed.
     
  4. Dec 27, 2004 #3
    Assume the conditions are independent.

    P.S. In order to win this case, BOTH conditions must occur.
     
  5. Dec 27, 2004 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    P(1 and two occur) is the product of the probability 1 occurs and the probability 2 occurs , as surely any introduction to probability theory states.
     
  6. Dec 27, 2004 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If events A and B are independent then the probability that A and B both occur is P(A and B)= P(A)*P(B). If the probability that A will occur is 0.70 and the probability that B will occur is 0.60 then the probability that they will both occur is (0.70)*(0.60)= 0.42.
     
  7. Dec 27, 2004 #6
    Matt/Halls

    Thank you sirs! That was most helpful. I've always kind of wondered how to calculate such things. Well, I'm off to go calculate the chances of me winning the lottery.
     
  8. Dec 27, 2004 #7
    Well actually, if anyone is feeling bored, could anyone explain to me why it is the case that the possibility of 2 independent conditions occuring simultaneously is the product of the possibility of occurence of each of the independent conditions. Or if that would take to long, could you direct me to a site which details proof of this. Thx.
     
  9. Dec 27, 2004 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Check this out.
    site 1
    site 2
    Apparently it is taken "a priori",as a postulate.Theory of probabilities has it's own fundamental laws.This is one of them.

    Daniel.
     
  10. Dec 27, 2004 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    An easy way of seeing this, is the following argument:
    1) Consider throwing two dice.
    Clearly, the probability of either die turning up a 6 can't influence the other die's probability of getting 6.
    That's basically what's meant with "independent" probabilities.
    (Of course, the probability of getting 2 6's can't differ if you throw the two dice simultaneously, or in sequence, either)
    2)
    Now, the probability of a "fair" die in getting "6" is clearly 1/6, since there exist 6 possible outcomes ("6" being one of those outcomes)
    3) But this means that the probability of getting 2 6's must be 1/6*1/6=1/36

    This result makes sense in terms of 2), since there are 6*6=36 possible outcomes of throwing two dice, and a double six is a single possibility among those outcomes.
     
  11. Dec 27, 2004 #10
    Here's a way of thinking about it.

    Let's say you have 100 dollars, and you are going to do a few activities that cost different amounts of money.

    Now, instead of fixed prices, the costs are percentages of how much money you have left. By some cosmic coincidence, the price of an activity is equal to the chance of it NOT happeneing (weird pricing system, I know).

    So, you do activity #1 successfully. This costs you 40% of your money (since it had a 60% chance of happening). Since you started with 100 dollars, you have to pay 40 dollars, and now have 60 dollars left.

    Feeling rich, you do activity number two successfully. This had a 70% chance of success, so it costs you 30% of your money. You have 60 dollars left. 30% of 60 is 18, so you have to pay 18 dollars, leaving you with 42 dollars.

    Since these are magic percentile dollars, that actually means you have 42% chance of getting to that point (i.e. both events happening).
     
  12. Dec 27, 2004 #11

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    "Well actually, if anyone is feeling bored, could anyone explain to me why it is the case that the possibility of 2 independent conditions occuring simultaneously is the product of the possibility of occurence of each of the independent conditions. Or if that would take to long, could you direct me to a site which details proof of this. Thx."


    Hmm, yes, ok, but I think you'll find that it's the definition of independent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Percentage Question
  1. Calculate percentages (Replies: 1)

  2. Percentage help (Replies: 7)

  3. Multiple Percentages (Replies: 2)

  4. Percentage reduction (Replies: 10)

Loading...