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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.

Answer :

lublana

Answer with Step-by-step explanation:

We are given that

[tex]P(R_1)=0.05[/tex]

[tex]P(R_2)=0.6[/tex]

[tex]P(R_3)=0.35[/tex]

[tex]P(Q/R_1)=0.40[/tex]

[tex]P(Q/R_2)=0.30[/tex]

[tex]P(Q/R_3)=0.60[/tex]

[tex]P(Q)=P(R_1)P(Q/R_1)+P(R_2)P(Q/R_2)+P(R_3)P(q/R_3)[/tex]

Substitute the values in the formula

a.[tex]P(Q)=0.05(0.40)+0.6(0.30)+0.35(0.60)=0.41[/tex]

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

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

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

[tex]P(R_2\cap Q)=P(R_2)\cdot P(Q/R_2)[/tex]

b.[tex]P(R_2\cap Q)=0.6(0.30)=0.18[/tex]

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

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

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