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

To determine the probability that the blind man picks a green bag, we will consider the distribution of bags in each room and use probability rules to find the solution. 

Let's identify the rooms:

• Room 1: Contains 3 blue, 4 green, and 5 red bags. Thus, there are a total of \(3 + 4 + 5 = 12\) bags in this room.
• Room 2: Contains 2 blue, 1 green, and 3 red bags. Thus, there are a total of \(2 + 1 + 3 = 6\) bags in this room.

Step 1: Calculate the probability of picking a green bag from each room.

• Probability of picking a green bag from Room 1: \(\frac{4}{12} = \frac{1}{3}\).
• Probability of picking a green bag from Room 2: \(\frac{1}{6}\).

Step 2: Calculate the probability of entering each room. Since the man chooses a room randomly, the probability of entering either room is \(\frac{1}{2}\).

Step 3: Applying the law of total probability to find the probability of picking a green bag.

The total probability of picking a green bag is given by:

\(P(\text{Green Bag}) = \left(\frac{1}{2} \right) \cdot \left(\frac{1}{3} \right) + \left(\frac{1}{2} \right) \cdot \left(\frac{1}{6} \right)\)

This simplifies to:

\(= \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{6} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\)

Hence, the probability that the blind man picks a green bag is \(\frac{1}{4}\).

Therefore, the correct answer is: \(\frac{1}{4}\).