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Math Help - Probability Question

  1. #1
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    Probability Question

    5 balls are picked at random one at a time and then replaced. If 65% of balls are blue, 30% are red and 5% are green. Then what is the probability that
    i) All 5 balls are red
    ii) 3 are blue, 1 is green and 1 is red
    iii) At least 1 is a green
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  2. #2
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    Quote Originally Posted by feage7 View Post
    5 balls are picked at random one at a time and then replaced. If 65% of balls are blue, 30% are red and 5% are green. Then what is the probability that
    i) All 5 balls are red
    ii) 3 are blue, 1 is green and 1 is red
    iii) At least 1 is a green
    BTW, if the balls were not replaced, then this problem would be unsolvable due to the unknown quantity of balls.

    i) P(red)^5 = 0.3^5
    ii) P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05
    iii) P(G \ge 1) = 1 - P(ge = 0) = 1 - 0.95^5
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  3. #3
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    Quote Originally Posted by colby2152 View Post
    BTW, if the balls were not replaced, then this problem would be unsolvable due to the unknown quantity of balls.

    i) P(red)^5 = 0.3^5
    ii) P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05 Mr F says: I think this answer might need to be multiplied by {\color{red}\frac{5!}{3! \, 1! \, 1!}}.
    iii) P(G \ge 1) = 1 - P(ge = 0) = 1 - 0.95^5
    To the OP: Read Multinomial distribution - Wikipedia, the free encyclopedia
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  4. #4
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    Mr. F, I don't understand why you multiplied it by 20?
    because if you pick, that's a combination, so the arrangement shouldn't matter, I might be wrong, since I am new to probability
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  5. #5
    Pim
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    The reason it has to be multiplied is exactly because order does not matter.
    <br />
P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05<br />

    Assumes that you first pick three blue balls, then a green one, then a red one. Because order does not matter, you'll need to multiply this by the amount of combinations possible, which in this case is:
    <br />
{\color{red}\frac{5!}{3! \, 1! \, 1!}}
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