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Math Help - Law of Total Probability

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    Law of Total Probability

    One bag contains 6 white and 4 black balls, and a second bag contains 3 white and 7 black balls. One ball is drawn at random from the first bag and placed unseen in the second bag.
    (a) Find the probability that a ball now drawn at random from the second bag will be black.
    (b) If the ball drawn from the second bag is black, find the probability that a black ball was transferred from the first bag.

    I drew a tree diagram for both of them but the fact that the balls get switched into another bag confuses me : / need some help
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  2. #2
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    Quote Originally Posted by Link88 View Post
    One bag contains 6 white and 4 black balls, and a second bag contains 3 white and 7 black balls. One ball is drawn at random from the first bag and placed unseen in the second bag.
    (a) Find the probability that a ball now drawn at random from the second bag will be black.
    (b) If the ball drawn from the second bag is black, find the probability that a black ball was transferred from the first bag.

    I drew a tree diagram for both of them but the fact that the balls get switched into another bag confuses me : / need some help
    Hi Link88,

    (a) there are 2 cases

    1. a white ball was transferred from the 1st bag and a black ball was chosen from the 2nd..
    this probability is

    \frac{6}{10}\frac{7}{11}

    since there are now 11 balls in the 2nd bag.

    2. a black ball was transferred from the 1st bag and a black ball was chosen from the 2nd...
    this probability is

    \frac{4}{10}\frac{8}{11}

    Summing these probabilities gives the probability of choosing a black from
    the 2nd bag after a ball was transferred from the 1st.

    (b) The probability that a black ball was moved to the 2nd bag is 0.4.

    Hence, maybe this question should read...

    if a black ball is chosen, what is the probability that it came from the 1st bag?

    It has no chance of being from the 1st bag if a white was transferred.

    It has a \frac{1}{8} chance of being from the 1st bag if a black was transferred.

    Hence, the probability is

    (0)\frac{1}{7}+\frac{4}{10}\frac{1}{8}
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