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  1. #1
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    probability

    There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by allrighty View Post
    There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
    find the probabilities for each case and add them up

    by (all possible) cases, i mean:

    probability of choosing 1 white, 0 black, 1 blue, 1 red, 3 green

    probability of choosing 1 white, 0 black, 0 blue, 2 red, 3 green

    probability of choosing 1 white, 0 black, 2 blue, 0 red, 3 green

    .
    .
    .
    .

    or we could do:

    1 - probability of choosing no white and no black ball

    so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors. there is a formula for that sort of thing. can't remember right now, i'll have to look it up. but when you do get it, just multiply the answer by 0.2 and that will give you the probability of choosing no white and no black ball
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors
    Is that right? A white AND a black are required.

    I have another method but it's messy.

    You could have for example 2W, 1B and 3 others 6!/(3!2!) ways each with probability 0.2^3 0.6^3

    or 5W, 1B 6!/5! ways with probability 0.2^6

    Having said all that, I'm a bit rubbish at probability so I'll wait and see what people say.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by a tutor View Post
    Is that right? A white AND a black are required.
    yes, i believe so. what i said was i want to find the probability of having 0 white AND 0 black. and then take 1 minus that probability to find the probability of at least 1 white or 1 black. i think what i described does the trick... but then again, i'm a noob when it comes to probability as well
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  5. #5
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    Original question said..

    Quote Originally Posted by allrighty View Post
    What is the probability that, from the 6 balls picked, there are white balls and black balls?
    and you said..

    Quote Originally Posted by Jhevon View Post
    to find the probability of at least 1 white or 1 black.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by a tutor View Post
    Original question said..



    and you said..



    ah yes. my bad. i didn't see the "and." i was under the impression if we have either or we were good...
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  7. #7
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    Hello, allrighty!

    I think I've solved it . . .


    There are 5 different colors of balls: white, black, blue, red, green.
    We randomly pick 6 balls.
    Each ball has a probability of 0.2 of getting each of the 5 colors.
    What is the probability that, from the 6 balls picked, there are white balls and black balls?
    The opposite of "some White and some Black" is "no White or no Black".


    To get no White, we must pick six balls from the other four colors.
    . . Then: . P(\text{0 White}) \:=\:(0.8)^6

    To get no Black, we must pick six balls from the other four colors.
    . . Then: . P(\text{0 Black}) \:=\:(0.8)^6

    To get no White and no Black, we pick six balls from the other 3 colors.
    . . Then: . P(\text{0 White} \,\wedge \,\text{0 Black}) \:=\;(0.6)^6


    Hence: . P(\text{0 White} \,\vee \,\text{0 Black}) \;=\;P(\text{0 White}) + P(\text{0 Black}) - P(\text{0 White} \,\wedge \,\text{0 Black})
    . . . . . . . . . . . . . . . . . . . . = \quad\;\;(0.8)^6\quad \;+ \;\quad(0.8)^6 \qquad- \qquad(0.6)^6
    . . . . . . . . . . . . . . . . . . . . = \qquad 0.4777632


    Therefore: . P(\text{some White and some Black}) \;=\;1 - 0.4777632 \;=\;\boxed{\:0.522368\:}

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  8. #8
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    A neat solution Soroban.

    I got the same answer rather more clumsily.

    I wrote a quick easy program to do it the way I mentioned above.
    Attached Thumbnails Attached Thumbnails probability-prob-prob.jpg  
    Last edited by a tutor; January 28th 2008 at 02:25 AM.
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