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Math Help - Conditional Probability #2

  1. #1
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    Conditional Probability #2

    A store has two gumball machines, which are stocked with maroon and gold gumballs. One machine has 100 maroon gumballs and 150 gold gumballs. The other has 150 maroon gumballs and 200 gold gumballs. THe owner has found out that customers are buying gu m from the first machine 40% of the time and from the second machine 60% of the time.

    a. What is the probability that a customer gets a maroon gumball?

    b. If a customer gets a gold gumball, what is the probability that it came from the first machine?


    What I got...

    a. P(maroon) = (250 maroon gumballs / 600 total gumballs) = 5/12

    I'm not too sure because of the 40% and 60%. Does that play a factor to the problem?



    b. We learned about this formula:

     <br />
P(A|B) = \frac{P(B | A)\, P(A)}{P(B|A) P(A) + P(B|A^C) P(A^C)}. \!<br />

    I suppose I would need to know if part a is right before I can continue.
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  2. #2
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    Quote Originally Posted by toop View Post
    A store has two gumball machines, which are stocked with maroon and gold gumballs. One machine has 100 maroon gumballs and 150 gold gumballs. The other has 150 maroon gumballs and 200 gold gumballs. THe owner has found out that customers are buying gu m from the first machine 40% of the time and from the second machine 60% of the time.

    a. What is the probability that a customer gets a maroon gumball?

    b. If a customer gets a gold gumball, what is the probability that it came from the first machine?


    What I got...

    a. P(maroon) = (250 maroon gumballs / 600 total gumballs) = 5/12

    I'm not too sure because of the 40% and 60%. Does that play a factor to the problem? Mr F says: Yes.

    [snip]
    My advice is to draw a tree diagram. Then it should be clear that the answer to (a) is (0.6)(100/250) + (0.4)(150/350) = ....
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  3. #3
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    Hello, toop!

    A store has two gumball machines, which are stocked with maroon and gold gumballs.
    Machine A has 100 maroon gumballs and 150 gold gumballs.
    Machine B has 150 maroon gumballs and 200 gold gumballs.
    The owner has learned that customers are buying from Machine A 40% of the time
    and from Machine B 60% of the time.

    (a) What is the probability that a customer gets a maroon gumball?
    You can't "dump them together" like that . . .


    P(\text{machine A}) = \tfrac{2}{5}\qquad \begin{array}{ccccc}P(m) &=& \tfrac{100}{250} &=& \tfrac{2}{5} \\ \\[-4mm]<br />
P(g) &=& \tfrac{150}{250} &=& \tfrac{3}{5}<br />
\end{array}

    . . Hence: . \begin{array}{ccccc}P(A \cap m) &=& \frac{2}{5}\cdot\frac{2}{5} &=& \frac{4}{25} \\ \\[-4mm]<br />
P(A \cap g) &=& \frac{2}{5}\cdot\frac{3}{5} &=& \frac{6}{25} \end{array}


    P(\text{machine B}) = \tfrac{3}{5}\qquad\begin{array}{ccccc}<br />
P(m) &=& \frac{150}{350} &=& \frac{3}{7} \\  \\[-4mm] P(g) &=&\frac{200}{350} &=& \frac{4}{7}\end{array}

    . . Hence: . \begin{array}{ccccc}P(B \cap m) &=& \frac{3}{5}\cdot\frac{3}{7} &=& \frac{9}{35} \\ \\[-4mm]<br />
P(B \cap g) &=& \frac{3}{5}\cdot\frac{4}{7} &=& \frac{12}{35}\end{array}


    Therefore: . P(m) \;=\;P(A \cap m) + P(B \cap m) \;=\;\frac{4}{25} + \frac{9}{35} \;=\;\frac{73}{175}




    (b) If a customer gets a gold gumball, what is the probability that it came from Machine A?
    I would use Bayes' Theorem: . P(A|B) \;=\;\frac{P(A \cap B)}{P(B)}

    We have: . P(A|g) \;=\;\frac{P(A \cap g)}{P(G)} .[1]

    . . We know: . P(A \cap g) \:=\:\frac{6}{25}
    . . And: . P(g) \;=\;P(A \cap g) + P(B \cap g) \;=\;\frac{6}{25} + \frac{12}{35} \;=\;\frac{102}{175}

    Substitute into [1]: . P(A|g) \;=\;\frac{\frac{6}{25}}{\frac{102}{175}} \;=\;\frac{7}{17}

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