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11 October, 20:23

Bowl B₁ contains two white chips, bowl B₂ contains two red chips, bowl B₃ contains two white chips and two red chips, and bowl B₄ contains three white chips and one red chip. The probabilities of selecting bowl B₁, B₂, B₃, or B₄ are 1/2, 1/4, 1/8 and 1/8, 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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  1. 11 October, 22:08
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    1) The probability of selecting a white chip is = 21/32

    2) The conditional probability that bowl B₁ had been selected, given that a white chip was drawn = 16/21

    Step-by-step explanation:

    Let

    B₁ = The event of randomly selecting Bowl B₁;

    B₂ = The event of randomly selecting Bowl B₂;

    B₃ = The event of randomly selecting Bowl B₃ and

    B₄ = The event of randomly selecting Bowl B₄.

    The probability of selecting each of the four bowls are as follows, P (B₁) = 1/2, P (B₂) = 1/4. P (B₃) = 1/8, P (B₄) = 1/8

    Let

    W = The event of randomly selecting a white chip.

    The probability that a white chip is selected from a bowl is given as

    P (W | B₁) = 1 for bowl B₁

    P (W | B₂) = 0 for bowl B₂

    P (W | B₃) = 1/2 for bowl B₃

    P (W | B₄) = 3/4 for bowl B₄

    There are four ways of selecting a white chip: (1) selecting a white chip from Bowl B₁; or (2) selecting a white chip from Bowl B₂; or (3) selecting a white chip from Bowl B₃ or (4) selecting a white chip from Bowl B₄. That is, the probability that a white chip is selected is:

    P (W) = P[ (W∩B₁) ∪ (W∩B₂) ∪ (W∩B₃) ∪ (W∩B₄) ]



    Given that the events W∩B₁, W∩B₂, W∩B₃ and W∩ B₄ are mutually exclusive, and by Multiplication Rule, we have:

    P (W) = P (W| B₁) P (B₁) + P (W| B₂) P (B₂) + P (W|B₃) P (B₃) + P (W|B₄) P (B₄)

    Substituting the numbers from above

    P (W) = (1*1/2) + (0*1/4) + (1/2*1/8) + (3/4*1/8) = 1/2+0+1/16+3/32

    =21/32

    The probability of selecting a white chip is = 21/32

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

    Solution.

    We are interested in finding P (B₁| W). We will use the fact that P (W) = 21/32, as seen from above in our previous calculation:

    From conditional probability P (B₁|W) = P (B₁∩W) / (P (W)) = and from multiplication Rule P (W|B₁) * P (B₁) / (P (W)) = 1 * (1/2) : (21/32) = 16/21

    Where P (W|B₁) = 1

    P (B₁ Given W), the conditional probability that bowl B₁ had been selected, given that a white chip was drawn = 16/21
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