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Math Help - odds of independent events

  1. #1
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    odds of independent events

    Forgive me, I read through that entire sticky post at the top, and at the end, if anything, I was more confused. So I'll spell everything out here, and if one of you fine talented folk can tell me if I'm on the right track or not you will make my day!

    I have ten cards turned face down. One has a square on it, the other 9 have circles.

    I believe I have a 1/10 or 10% chance of drawing the square.
    I believe that I have a 9/10 or 90% chance of drawing a circle card.


    If, after each draw, the cards are shuffled, so that it is an entirely random process once again, entirely separate from the initial draw, what are the odds that after two draws (at least) one of them would have been a square?

    1/10 * 1/10 = 1/100? But that's not right. I know the odds are higher than that.

    9/10 * 9/10 = 81/100 of having drawn nothing but circles? That sounds like it could be right, and if so means that the remainder, 19% is the correct odds for (at least) 1 of my 2 cards having been a square?

    9/10 * 9/10 * 9/10 = 729/1000 for 27.1% after 3.

    9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 387420489 / 1000000000 = 0.387420489 or ~61.25% of after 9 draws having (at least) 1 of them be a square.

    9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 3486784401 / 10000000000 = 0.3486784401 or ~65.1% after 10 draws

    The first part of my question is, am I right?


    The second part of my question is:

    How many times would I need to draw to have a 90% chance of having drawn at least 1 square card?
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  2. #2
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    Re: odds of independent events

    Quote Originally Posted by BarryAllen View Post
    Forgive me, I read through that entire sticky post at the top, and at the end, if anything, I was more confused.
    Well, all that really says that you show us what you DO know and what you can do so that we can give hints and suggestions without doing the problem for you.

    So I'll spell everything out here, and if one of you fine talented folk can tell me if I'm on the right track or not you will make my day!

    I have ten cards turned face down. One has a square on it, the other 9 have circles.

    I believe I have a 1/10 or 10% chance of drawing the square.
    I believe that I have a 9/10 or 90% chance of drawing a circle card.
    Yes, those are the correct values for one draw.


    If, after each draw, the cards are shuffled, so that it is an entirely random process once again, entirely separate from the initial draw, what are the odds that after two draws (at least) one of them would have been a square?
    That is, you draw a card, look at it, put the card back, then shuffle and draw another. This is called "sampling with replacement".

    1/10 * 1/10 = 1/100? But that's not right. I know the odds are higher than that.
    No, it is not correct. Since 1/10 is the probability of getting the square, 1/10* 1/10 is the probability that both cards have the square, not just "at least one". You also have to consider the probability that the first card has a square (1/10) and the second card does NOT (9/10). What is the probablity of that? And you have to consider the probability that the first cards does NOT have the square but the second card does not. What is the probability of that?

    As a check, if there is a 9/10 chance that a card does not have a square, what is the chance that two do not? Subtract that from 1 to get the probability that NOT both cards do not have the square (which is the same as "at least one does have the square).

    9/10 * 9/10 = 81/100 of having drawn nothing but circles? That sounds like it could be right, and if so means that the remainder, 19% is the correct odds for (at least) 1 of my 2 cards having been a square?
    Yes, that is exactly correct.

    9/10 * 9/10 * 9/10 = 729/1000 for 27.1% after 3.

    9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 387420489 / 1000000000 = 0.387420489 or ~61.25% of after 9 draws having (at least) 1 of them be a square.

    9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 3486784401 / 10000000000 = 0.3486784401 or ~65.1% after 10 draws

    The first part of my question is, am I right?
    Those are the probablities you do NOT get the square in that number of draws. Of course the probability of getting the square on the first draw is 1/10, on the first or second draws is 1/10+ (9/10)(1/10)= (1/10)(1+ 9/10), on the first second or third draws (1/10)(1+ 9/10+ 9^2/10^2), on the first, second, third, or fourth draws is (1/10)(1+ 9/10+ (9/10)^2+ (9/10)^3). Can you see what you get on the "n"th draw? Think "geometric series".


    The second part of my question is:

    How many times would I need to draw to have a 90% chance of having drawn at least 1 square card?
    So you want (1/10)(1+ 9/10+ (9/10)^2+ (9/10)^3+ ...)= 9/10. Again, think "geometric series".
    Last edited by HallsofIvy; December 26th 2012 at 06:50 PM.
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