An urn contains 20 red balls 30 blue and the rest green 40 o

An urn contains 20% red balls, 30% blue, and the rest green. 40% of the red balls, 30% of the blue balls, and 10% red balls are defective. (is this possible? 40% + 30% + 10%

Solution

Let R shows the event that ball is red, B shows the event that ball is blue and G shows the event that the ball is green. So we have

P(R)=0.20

P(B)=0.30

P(G)=0.50

Let D shows the event that ball is defective. Since 40% of the red ball are defective so we have

P(D|R)=0.40

Since 30% of the blue ball are defective so we have

P(D|B)=0.30

And since 10% of the green ball are defective so we have

P(D|G)=0.10

(c)

So

P(D)=P(D|R)*P(R)+P(D|B)*P(B)+P(D|G)*P(G)=0.20*0.40+0.30*0.30+0.50*0.10=0.08+0.09+0.05=0.22

Hence, the probability that ball is defective is 0.22

(d)

There is no Urn III so the probability that ball came from Urn III is 0.