Question

A blind man lives in an apartment containing 2 rooms. Each day before going to work he enters any one room randomly, picks up a bag and leaves home. One of the rooms contains 3 blue, 4 green and 5 red bags and the other contains 2 blue, 1 green and 3 red bags. What is the probability that he takes a green bag to his workplace?

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{2}{3}\)
• E. \(\frac{3}{4}\)

Answer 1

The Correct Option is A

The correct answer is option (A):

\(\frac{1}{4}\)

Here's how to solve this probability problem step-by-step:

First, calculate the probability of the blind man selecting each room. Since he chooses randomly, the probability of selecting room 1 is 1/2, and the probability of selecting room 2 is also 1/2.

Next, find the probability of selecting a green bag from each room.

Room 1: There are 4 green bags and a total of 3 + 4 + 5 = 12 bags. The probability of picking a green bag from room 1 is 4/12 = 1/3.

Room 2: There is 1 green bag and a total of 2 + 1 + 3 = 6 bags. The probability of picking a green bag from room 2 is 1/6.

Now, use the law of total probability. This states that the probability of an event (picking a green bag) is the sum of the probabilities of that event occurring in each scenario (room) weighted by the probability of each scenario.

Probability (Green Bag) = [Probability (Room 1) * Probability (Green Bag | Room 1)] + [Probability (Room 2) * Probability (Green Bag | Room 2)]

Probability (Green Bag) = (1/2 * 1/3) + (1/2 * 1/6)
Probability (Green Bag) = 1/6 + 1/12
Probability (Green Bag) = 2/12 + 1/12
Probability (Green Bag) = 3/12 = 1/4

Therefore, the probability that the blind man takes a green bag is 1/4.