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  1. #1
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    probability

    1) A school offers japanese and arabic as two foreign languages. the probabilitiy that a student who studies japanese also studies arabic is 1/4 and 1/5 of the students who study arabic also studies japanese too. among the students, 1/5 of them do not study any of the two languages . what is the probability that a student in the school studies both languages? Are the events of studying japanese and arabic independent? ( answer provided 1/10,not independent) explain pls

    2)Bag A contains 3 red balls and 2 green balls whereas bag B contain 2 red balls and 3 green balls. two balls r withdrwan from bag a to be put into bag b . then two balls are taken out from B to be put in bag A. what is the probability that bag A now contains 3 red balls and 2 green balls.? answer 10/21
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    Quote Originally Posted by qweiop90 View Post
    1) A school offers japanese and arabic as two foreign languages. the probabilitiy that a student who studies japanese also studies arabic is 1/4 and 1/5 of the students who study arabic also studies japanese too. among the students, 1/5 of them do not study any of the two languages . what is the probability that a student in the school studies both languages? Are the events of studying japanese and arabic independent? ( answer provided 1/10,not independent) explain pls

    [snip]
    Draw a Karnaugh table:

    \begin{tabular}{l | c | c | c} & J & J$\, '$ & \\ \hline A & a & b & \\ \hline A$\, '$ & c & 1/5 & \\ \hline & & & 1 \\ \end{tabular}<br />

    From the given data you know that:

    \frac{a}{a+b} = \frac{1}{5} \Rightarrow b = 4a .... (1)

    \frac{a}{a+c} = \frac{1}{4} \Rightarrow c = 3a .... (2)

    a + b + c = \frac{4}{5}.

    Solve for a, b and c.

    From the Karnaugh table \Pr (A \cap J) = a.

    A and J are independent if \Pr (A \cap J) = \Pr(J) \cdot \Pr(A).

    \Pr(J) = a + c = \, ....
    \Pr(A) = a + b = ....

    Does \Pr (A \cap J) = \Pr(J) \cdot \Pr(A) ....?
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  3. #3
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    Quote Originally Posted by qweiop90 View Post
    [snip]
    2)Bag A contains 3 red balls and 2 green balls whereas bag B contain 2 red balls and 3 green balls. two balls r withdrwan from bag a to be put into bag b . then two balls are taken out from B to be put in bag A. what is the probability that bag A now contains 3 red balls and 2 green balls.? answer 10/21
    Two balls taken from bag A. The possibilities are:

    2R => Bag B has 4R and 3G. Probability = (3/5)(2/4) = 6/20. Bag A has 1R, 2G.

    1R, 1G => Bag B has 3R and 4G. Probability = 2 (3/5)(2/4) = 12/20. Bag A has 2R, 1G.

    2G => Bag B has 2R and 5G. Probability = (2/5)(1/4) = 2/20. Bag A has 3R.


    Now consider each of the possible bag B's:

    4R, 3G: The two balls taken need to be 2R for Bag A to have 3R and 2G. The probability of this is (4/7)(3/6) = 12/42.

    3R, 4G: The two balls taken need to be 1R 1G for Bag A to have 3R and 2G. The probability of this is 2 (3/7)(4/6) = 24/42.

    2R 5G: The two balls taken need to be 2G for Bag A to have 3R and 2G. The probability of this is (5/7)(4/6) = 20/42.


    So the total probability is (6/20)(12/42) + (12/20)(24/42) + (2/20)(20/42) = .......
    Last edited by mr fantastic; August 19th 2008 at 09:18 PM. Reason: Added a bit more spacing
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