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. \(\dfrac{1}{4}\)
• B. \(\dfrac{1}{3}\)
• C. \(\dfrac{1}{2}\)
• D. \(\dfrac{2}{3}\)

Hint:

Apply the law of total probability over the two equally likely rooms.

Answer 1

The Correct Option is A

Step 1: Find the chance of a green bag in each room.
The first room has \(3+4+5=12\) bags in total, with 4 green, so \(P(\text{green} \mid \text{room 1}) = \dfrac{4}{12} = \dfrac{1}{3}\).
The second room has \(2+1+3=6\) bags in total, with 1 green, so \(P(\text{green} \mid \text{room 2}) = \dfrac{1}{6}\).

Step 2: Account for the random room choice.
The man enters either room with equal chance, so \(P(\text{room 1}) = P(\text{room 2}) = \dfrac{1}{2}\).

Step 3: Combine using the law of total probability.
\(P(\text{green}) = P(\text{room 1}) \times P(\text{green} \mid \text{room 1}) + P(\text{room 2}) \times P(\text{green} \mid \text{room 2})\).
\(P(\text{green}) = \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{6} + \dfrac{1}{12}\).

Step 4: Add the fractions.
\(\dfrac{1}{6} + \dfrac{1}{12} = \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12} = \dfrac{1}{4}\).

Final Answer:
The probability of taking a green bag is \(\dfrac{1}{4}\). \[ \boxed{\dfrac{1}{4}} \]