” For mutually exclusive events R1, R2, and R3, we have P(R1) = 0.05, P(R2) = 0.6, and P(R3) = 0.35. Also, P(Q|R1) = 0.40, P(Q|R2) = 0.30, a

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” For mutually exclusive events R1, R2, and R3, we have P(R1) = 0.05, P(R2) = 0.6, and P(R3) = 0.35. Also, P(Q|R1) = 0.40, P(Q|R2) = 0.30, and P(Q|R3) = 0.60. Find the following. (a) P(R1|Q) (b) P(R2|Q)” Excerpt From: Erin N. Bodine. “Mathematics for the Life Sciences.” Apple Books.

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Sophia 2 weeks 2021-09-12T09:58:45+00:00 1 Answer 0

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    2021-09-12T09:59:47+00:00

    Answer with Step-by-step explanation:

    We are given that

    P(R_1)=0.05

    P(R_2)=0.6

    P(R_3)=0.35

    P(Q/R_1)=0.40

    P(Q/R_2)=0.30

    P(Q/R_3)=0.60

    P(Q)=P(R_1)P(Q/R_1)+P(R_2)P(Q/R_2)+P(R_3)P(q/R_3)

    Substitute the values in the formula

    a.P(Q)=0.05(0.40)+0.6(0.30)+0.35(0.60)=0.41

    P(R_1\cap Q)=P(Q/R_1)P(R_1)=0.05\times 0.40=0.02

    P(R_1/Q)=\frac{P(R_1\cap Q)}{P(Q)}

    P(R_1/Q)=\frac{0.02}{0.41}=0.049

    P(R_2\cap Q)=P(R_2)\cdot P(Q/R_2)

    b.P(R_2\cap Q)=0.6(0.30)=0.18

    P(R_2/Q)=\frac{P(R_2\cap Q)}{P(Q)}

    P(R_2/Q)=\frac{0.18}{0.41}=0.439

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