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Math Help - Probability Questions

  1. #1
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    Probability Questions

    I always have disliked probablility...

    So... I need some help.. If you will please much appreciated..

    How many different signals can be made using four flags of different colors on a vertical flagpole if exactly three flags are used for each signal?



    How many 5 letter code words are possible in the word SKATE

    Thanks )
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  2. #2
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    Quote Originally Posted by skyslimit View Post
    I always have disliked probablility...

    So... I need some help.. If you will please much appreciated..

    How many different signals can be made using four flags of different colors on a vertical flagpole if exactly three flags are used for each signal?



    How many 5 letter code words are possible in the word SKATE

    Thanks )
    Q1. ^4 P_3 = (4)(3)(2) = 24 (since the order is important).

    Q2. 5! = ....
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  3. #3
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    OK...

    Also, if you will please check my answer on this one:

    If a box of soda contains 36 bottles, 8 are white, 5 are tan 6 are pink 1 is purple, 2 are yellow, 4 is orange, 10 are green

    Select 9 bottles randomly, probability that 3 are white

    My answer: 8 c 3 / 36 c 9 ? This is right or is there more to it...

    Harder problem..

    Probability that 3 are white, two are tan, one is pink, one is yellow, and and two are green
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  4. #4
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    Quote Originally Posted by skyslimit View Post
    OK...

    Also, if you will please check my answer on this one:

    If a box of soda contains 36 bottles, 8 are white, 5 are tan 6 are pink 1 is purple, 2 are yellow, 4 is orange, 10 are green

    Select 9 bottles randomly, probability that 3 are white

    My answer: 8 c 3 / 36 c 9 ? This is right or is there more to it... Mr F says: There's more to it - see below.

    Harder problem..

    Probability that 3 are white, two are tan, one is pink, one is yellow, and and two are green Mr F says: Please post the harder problem exctly as it's written.
    \frac{^8C_3 \cdot {\color{red}^{28}C_6}}{^{36}C_9}
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  5. #5
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    Harder problem
    If you select 9 bottles, without replacement, what it is probability that 3 will be white, 2 will be tan, 1 is pink, one is yellow, and two are green?

    And one last one after that... seems simple but quickly turns complicated,

    How many sundaes?

    Ice Cream Flavors

    Chocolate
    Cookies N Cream
    Strawberry
    Mint

    Toppings

    Nut
    Caramel
    Gummy Bears
    Oeros
    Fudge
    Butterscotch

    How many sundaes are possible when choosing 1 sundae and one topping?


    Help much appreciated... Again not my favorite subject to deal with in math
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  6. #6
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    Probability

    Hello skyslimit
    Quote Originally Posted by skyslimit View Post
    Harder problem
    If you select 9 bottles, without replacement, what it is probability that 3 will be white, 2 will be tan, 1 is pink, one is yellow, and two are green?
    The total number of ways of selecting 9 from 36 is \binom{36}{9}.

    You then have to choose:

    • 3 white from 8. This can be done in \binom83 ways.
    • 2 tan from 5. This can be done in \binom52 ways.
    • 1 pink from 6. This can be done in \binom?? ways


    • 1 yellow from 2 ...?


    • 2 green from 10 ...?

    When you have filled in the gaps in the last three, there are then two things that you need to do to complete the calculation:

    1. Multiply all 5 of these answers together to find the total number of ways of choosing all 5 colours.
    2. Divide this answer by \binom{36}{9} to find the probability that one of these selections occurs.

    And one last one after that... seems simple but quickly turns complicated,

    How many sundaes?

    Ice Cream Flavors

    Chocolate
    Cookies N Cream
    Strawberry
    Mint

    Toppings

    Nut
    Caramel
    Gummy Bears
    Oeros
    Fudge
    Butterscotch

    How many sundaes are possible when choosing 1 sundae and one topping?
    Use the same principle here. It's called the r-s principle, and it is:

    • If a task A can be carried out in r ways, and an independent task B can be carried out in s ways, then the number of ways in which both tasks can be carried out is r \times s.

    So:

    • How many ways are there of choosing an ice-cream flavour?
    • How many ways are there of choosing a topping?
    • Are these tasks independent of each other? (In other words, can any one of the ice-creams be combined with any of the toppings?) Answer: yes!

    Then use the r-s principle and multiply your two numbers together to find the total number of sundaes.

    If you want us to check your answers, post them here.

    Grandad
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    \frac{^8C_3 \cdot {\color{red}^{28}C_6}}{^{36}C_9}
    For this one, my question, the 28c6 is derived exactly from where? I have it as being being (28): white subtracted from total of 36, and 6, the leftover of selected bottles. Is this right?

    Also, for the harder one with multiple functions, do we not need to include what we have included previously?


    And for the sundae problem I worded it wrong. it is supposed to say, 1 flavor 3 toppings. Is 3 (choices) * 6 * 6 * 6 (toppings) right??
    Last edited by skyslimit; April 9th 2009 at 06:31 PM.
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  8. #8
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    Probability

    Hello skyslimit
    Quote Originally Posted by skyslimit View Post
    For this one, my question, the 28c6 is derived exactly from where? I have it as being being (28): white subtracted from total of 36, and 6, the leftover of selected bottles. Is this right?
    Correct!
    Also, for the harder one with multiple functions, do we not need to include what we have included previously?
    No. This is a separate question. So you begin again in the way that I explained.

    And for the sundae problem I worded it wrong. it is supposed to say, 1 flavor 3 toppings. Is 3 (choices) * 6 * 6 * 6 (toppings) right??
    No. First, there are 4 choices of flavour, not 3. Then, it depends upon whether the 3 toppings must all be different. If you're allowed to choose the same topping more than once, then the answer is 4 x 6 x 6 x 6. But if they must all be different, then it's 4 \times \binom{6}{3}.

    Grandad

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