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Math Help - Balls in an Urn

  1. #1
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    Balls in an Urn

    Twenty balls are placed in an urn. Five are red, five green, five yellow and five blue. Three balls are drawn from the urn at random without replacement. Write down expressions for
    the probabilities of the following events. (You need not calculate their numerical values.)

    a) Exactly one of the balls drawn is red.
    b) The three balls drawn have different colours.
    c) The number of blue balls drawn is strictly greater than the number of yellow balls
    drawn.
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  2. #2
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    Hello, Aquafina!

    Twenty balls are placed in an urn: 5 red, 5 green, 5 yellow, 5 blue.
    Three balls are drawn from the urn at random without replacement.
    There are: . {20\choose3} \,=\,1140 possible outcomes.


    Write expressions for the probabilities of the following events.
    (You need not calculate their numerical values.)

    a) Exactly one of the balls drawn is red.
    There are: 5 Reds and 15 Others.

    We want 1 Red and 2 Others. .There are: . {5\choose1}{15\choose2} \:=\:525 ways.

    . . Therefore: . P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}



    b) The three balls drawn have different colours.
    Select 3 of the 4 colors: . {4\choose3} \:=\:4 ways.
    To pick one of each of the colors, there are: . {5\choose1}{5\choose1}{5\choose1} \:=\:125 ways.

    Hence, there are: . 4\cdot125 \:=\:500 ways to get 3 colors.

    Therefore: . P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}



    c) The number of blue balls drawn is strictly greater than the number of yellow balls drawn.
    There are two cases to consider:

    . . [1] Two Blue and one Yellow: . {5\choose2}{5\choose1} \:=\:50 ways
    . . [2] Three Blue: . {5\choose3} \:=\:10 ways

    Hence, there are: . 50 + 10 \:=\:60 ways to have more Blue than Yellow.

    . . Therefore: . P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Aquafina!

    There are: . {20\choose3} \,=\,1140 possible outcomes.


    There are: 5 Reds and 15 Others.

    We want 1 Red and 2 Others. .There are: . {5\choose1}{15\choose2} \:=\:525 ways.

    . . Therefore: . P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}



    Select 3 of the 4 colors: . {4\choose3} \:=\:4 ways.
    To pick one of each of the colors, there are: . {5\choose1}{5\choose1}{5\choose1} \:=\:125 ways.

    Hence, there are: . 4\cdot125 \:=\:500 ways to get 3 colors.

    Therefore: . P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}



    There are two cases to consider:

    . . [1] Two Blue and one Yellow: . {5\choose2}{5\choose1} \:=\:50 ways
    . . [2] Three Blue: . {5\choose3} \:=\:10 ways

    Hence, there are: . 50 + 10 \:=\:60 ways to have more Blue than Yellow.

    . . Therefore: . P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}

    Hi I got the same answers for i) and ii) but not iii). First of all, can you explain your method in ii?

    I basically did:

    P(diff colours) = 5/20 * 5/19 * 5/18 * 3! * 4C3

    I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?

    for iii)

    I did P(1 blue, no yellow) as well

    Also, P(2 blue, no yellow)
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  4. #4
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    I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?
    You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:

    Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
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  5. #5
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    Quote Originally Posted by ANDS! View Post
    You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:

    Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
    Thanks.

    What about the next bit.

    for iii)

    I did P(1 blue, no yellow) as well

    Also, P(2 blue, no yellow)

    Why aren't these included in the answer given in the post above?
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  6. #6
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    Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
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  7. #7
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    Soroban only considered these 2 cases:

    Quote Originally Posted by Soroban View Post
    There are two cases to consider:

    . . [1] Two Blue and one Yellow: . {5\choose2}{5\choose1} \:=\:50 ways
    . . [2] Three Blue: . {5\choose3} \:=\:10 ways

    Hence, there are: . 50 + 10 \:=\:60 ways to have more Blue than Yellow.

    . . Therefore: . P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}

    Not the ones I mentioned as well. When I had worked the answer out, I included these 2 cases, along with the cases I mentioned to get the answer.

    I want to know why those are not needed.
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  8. #8
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    I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn).
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  9. #9
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    Quote Originally Posted by ANDS! View Post
    I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn).
    Thanks, whats your answer for that?

    I got 77/228
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