Bowl B1 contains one white and two red chips; bowl B2 contains two white and two red chips; bowl B3 contains one white and four red chips. T

Question

Bowl B1 contains one white and two red chips; bowl B2 contains two white and two red chips; bowl B3 contains one white and four red chips. The probabilities of selecting bowl B1, B2, or B3 are 0.3, 0.2, and 0.5, respectively. A bowl is selected using these probabilities and a chip is then drawn at random. Find:

a. P(W), the probability of drawing a white chip.
b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

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Diễm Kiều 5 years 2021-08-29T09:42:21+00:00 1 Answers 60 views 0

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    2021-08-29T09:44:13+00:00
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    Answer:

    a) 0.3 = 30% probability of drawing a white chip.

    b) 0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.

    Step-by-step explanation:

    Conditional Probability

    We use the conditional probability formula to solve this question. It is

    P(B|A) = \frac{P(A \cap B)}{P(A)}

    In which

    P(B|A) is the probability of event B happening, given that A happened.

    P(A \cap B) is the probability of both A and B happening.

    P(A) is the probability of A happening.

    a. P(W), the probability of drawing a white chip.

    1/3(one white out of 3) of 0.3(from B1).

    1/2(two white out of 4) of 0.2(from B2).

    1/5(one white out of 5) of 0.5(from B3). So

    P(W) = 0.3333*0.3 + 0.5*0.2 + 0.2*0.5 = 0.3

    0.3 = 30% probability of drawing a white chip.

    b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

    The probability of drawing a white chip from B1 is 1/3 out of 0.3, so:

    P(B1 \cap W) = 0.3\frac{1}{3} = 0.1

    Then the conditional probability is:

    P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.3} = 0.3333

    0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.

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