Question

A green urn contains three red balls and five black balls. A blue urn contains three red balls and seven black balls. A yellow urn contains two red balls and four black balls.

1. If an urn is selected at random and a ball is drawn, then the probability that it will be red is .
2. If a red ball is selected, then the probability that the green urn is selected is .
3. If the person drawing a ball from an urn favors choosing the yellow urn twice more than the green urn and thrice more than the blue urn, then the probability of drawing a red ball is .

Answer 1

Let A - ''red ball", H1 - "green urn", H2 - "blue urn", H3 - "yellow urn", then

1) According to the total probability formula$P(A)=P(H1)*P(A|H1)+P(H2)*P(A|H2)+P(H3)*P(A|H3)={\frac 1 3}*{\frac 3 8}+{\frac 1 3}*{\frac 3 {10}}+{\frac 1 3}*{\frac 2 6}={\frac {121} {360}}$

2) According to Bayesian formula

$P(H1|A)={\frac {P(H1)*P(A|H1)} {P(A)}}={\frac {\frac 1 8} {\frac {121} {360}}}={\frac {45} {121}}$

3) $P(H1)={\frac 3 {11}}$ $P(H2)={\frac 2 {11}}$ $P(H3)={\frac 6 {11}}$

According to the total probability formula

$P(A)=P(H1)*P(A|H1)+P(H2)*P(A|H2)+P(H3)*P(A|H3)={\frac 3 {11}}*{\frac 3 8}+{\frac 2 {11}}*{\frac 3 {10}}+{\frac 6 {11}}*{\frac 2 6}\approx 0.339$