Question

Three boxes contain red and green balls. Box 1 has 5 red balls and 5 green balls, Box 2 has 7 red balls and 3 green balls, and Box 3 contains 6 red balls and 4 green balls. The probability of choosing Box 1 is 1/4, Box 2 is 1/6, and Box 3 is 1/8. What is the probability that the ball chosen is green? What is the probability that the ball chosen is red

Answer 1

Let event A₁ = box 1 chosen,

event A₂ = box 2 chosen,

event A₃ = box 3 chosen.

Since sum of probabilities

$$P(A_1) + P(A_2) + P(A_3) = \frac{1}{4} + \frac{1}{6} + \frac{1}{8}=\frac{13}{24}\not =1$$

events A₁, A₂, A₃ are not exhaustive and Law of total probability can't be applied, and the problem should be modified.

For example we can find conditional probability that the ball chosen is green, given that we selected one of the boxes.

Let event G = selected ball is green,

event R = selected ball is red,

then conditional probability

$$P(G|(A_1\cup A_2\cup A_3))=$$

$$P(G\cap(A_1\cup A_2\cup A_3))/P(A_1\cup A_2\cup A_3)=$$

$$P((G\cap A_1)\cup (G\cap A_2)\cup (G\cap A_3))/P(A_1\cup A_2\cup A_3)$$

Conditional probabilities

P(G|A₁) = 5/(5+5) = 5/10 = 1/2, P(G|A₂) = 3/(3+7)=3/10, P(G|A₃) = 4/(4+6) = 4/10 = 2/5,

therefor

$$P((G\cap A_1)\cup (G\cap A_2)\cup (G\cap A_3))/P(A_1\cup A_2\cup A_3)=$$

$$\frac{(P(A_1)P(G|A_1)+P(A_2)P(G|A_2)+P(A_3)P(G|A_3))}{P(A_1\cup A_2\cup A_3)}=$$

$$\frac{1/4\cdot5/10+1/6\cdot3/10+1/8\cdot4/10}{1/4+1/6+1/8}=$$

$$\frac{5/40+1/20+1/20}{13/24}=\frac{9/40}{13/24}=$$

$$\frac{9\cdot24}{13\cdot40}=\frac{27}{65}.$$

Conditional probability

$$P(R|(A_1\cup A_2\cup A_3))=1-\frac{27}{65}=\frac{38}{65}$$

Answer to the modified problem: conditional probability that the ball chosen is green, given that we selected one of the boxes is 27/65, conditional probability that the ball chosen is red, given that we selected one of the boxes is 38/65.