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Math Help - Cookie Question!!

  1. #1
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    Cookie Question!!

    Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
    i) What is the probability that Fred picks a plain cookie?
    ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by matty888 View Post
    Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
    i) What is the probability that Fred picks a plain cookie?
    ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?
    Bowl 1 has 50% of the cookies and so too does Bowl 2.

    Bowl 1 contains 40% choc chip and 60% plain.

    Bowl 2 contains 50% choc chip and 50% plain.

    Use Bayes' Rule. (Dont you just love this rule? )

    <br />
P(B_{1} | P) = \frac{ P(P | B_{1}) P(B_{1})} {P(P | B_{1}) P(B_{1}) + P(P | B_{2}) P(B_{2})}

    <br />
P(B_{1} | P) = \frac{ (0,6)(0,5) }{ (0,6)(0,5) + (0,5)(0,5) }

     P(B_{1} | P) = \frac{6}{11}
    Last edited by janvdl; April 29th 2008 at 09:36 AM.
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  3. #3
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    Thanks

    Thanks for the answer,would I be right by saying that the answer to part i) is
    11/20
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  4. #4
    Member Danshader's Avatar
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    yup thats correct
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by matty888 View Post
    Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
    i) What is the probability that Fred picks a plain cookie?
    ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?
    Quote Originally Posted by matty888 View Post
    Thanks for the answer,would I be right by saying that the answer to part i) is
    11/20
    Sorry, I only did part 2... My apologies.

    For part 1:

    P(Plain) = P(P | B_{1})P(B_{1}) + P(P | B_{2})P(B_{2})

    P(Plain) = 0,6(0,5) + 0,5(0,5)

    P(Plain) = \frac{11}{20}
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  6. #6
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    Hello, Matty!

    A slightly different approach . . .


    Suppose there are two bowls full of cookies.
    Bowl #1 has 20 chocolate chip and 30 plain cookies; Bowl #2 has 25 of each.

    Fred picks a bowl at random and then picks a cookie at random.

    a) What is the probability that Fred picks a plain cookie?
    P(\text{Bowl \#1} \:\wedge\text{ Plain}) \:=\:\left(\frac{1}{2} \right)\left(\frac{30}{50}\right) \:=\:\frac{3}{10}

    P(\text{Bowl \#2} \:\wedge\text{ Plain}) \:=\:\left(\frac{1}{2}\right)\left(\frac{25}{50}\r  ight) \:=\:\frac{1}{4}

    . . Therefore: . P(\text{Plain}) \;=\;\frac{3}{10} + \frac{1}{4} \;=\;\boxed{\frac{11}{20}}




    bi) If the cookie Fred picked is a plain one,
    what is the probability that it was from Bowl #1?

    Bayes' Theorem: . P(A\,|\,B) \;=\;\frac{P(A\,\wedge B)}{P(B)}

    We have: . P(\text{Bowl \#1 }|\text{ Plain}) \;=\;\frac{P(\text{Bowl \#1 }\wedge\text{ Plain})}{P(\text{Plain})} \;=\;\frac{\dfrac{3}{10}}{\dfrac{11}{20}} \;=\;\boxed{\frac{6}{11}}

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  7. #7
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    Smile

    Hello and thanks vey much Soroban.Thanks janvdl and Danshader too!!!
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