Question:medium

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?

Show Hint

Apply the law of total probability over the two equally likely rooms.
Updated On: Jul 21, 2026
  • \(\dfrac{1}{4}\)
  • \(\dfrac{1}{3}\)
  • \(\dfrac{1}{2}\)
  • \(\dfrac{2}{3}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Convert each room's green share to a percentage.
Room 1 holds 12 bags with 4 green, which is about 33.3 percent green.
Room 2 holds 6 bags with 1 green, which is about 16.7 percent green.

Step 2: Weight each room by the 50 percent chance of entering it.
Room 1 contributes \(50\% \times 33.3\% = 16.65\%\) toward the overall chance of a green bag.
Room 2 contributes \(50\% \times 16.7\% = 8.35\%\) toward the overall chance.

Step 3: Add the two contributions.
The total chance is \(16.65\% + 8.35\% = 25\%\), since the rounding cancels out exactly.

Step 4: Convert back to a fraction.
25 percent equals \(\dfrac{25}{100} = \dfrac{1}{4}\), which matches the fraction obtained through direct probability rules.

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