Question

a) One of two boxes contain 4 red balls and 2 green balls and the second box contains 4 green and two red balls. By design, the probabilities of selecting box 1 or box 2 at random are 1/3 for box 1 and 2/3 for box 2. A box is selected at random and a ball is selected at random from it. a) Given that the ball selected is red, what is the probability it was selected from the first box? b) Given that the ball selected is red, what is the probability it was selected from the second box? c) Compare the results in parts a) and b) and explain the answer.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Let B₁ represent box 1, B₂ represent box 2, E₁ represent the event of selecting box 1, E₂ represent the event of selecting box 2 and R represent the event of selecting a red ball.

Given that:

Probability of selecting box 2 P(E₂) = 2/3, Probability of selecting box 1 = P(E₁) = 1/3

Probability of selecting red ball from box 1 = P(R | E₁) = 4/6 [4 red balls out of 4 red, 2 green]

Probability of selecting red ball from box 2 = P(R | E₂) = 2/6 [2 red balls out of 2 red, 4 green]

a)

Given the ball is red, the probability it was selected from the first box is:

$P(E_1|R)=\frac{P(R|E_1)P(E_1)}{P(R|E_1)P(E_1)+P(R|E_2)P(E_2)} =\frac{2/3*1/3}{(2/3*1/3)+(1/3*2/3))} =\frac{1}{2}$

b)

Given the ball is red, the probability it was selected from the second box is:

$P(E_2|R)=\frac{P(R|E_2)P(E_1)}{P(R|E_2)P(E_2)+P(R|E_1)P(E_1)} =\frac{1/3*2/3}{(1/3*2/3)+(2/3*1/3))} =\frac{1}{2}$

c)

We can see that both probabilities are equal. Although we have more red balls in box 1 (twice as much) than in box 2 but the probability of selecting from box 2 (is twice as much) than from selecting from box 1.