Q.31 Three balls, colored in blue, green and red, are successively transferred from box A
to box B in the order BLUE-GREEN-RED. The probability of a reverse transfer of the balls
to the box A in the same order (rounded off to two decimal places) is _________.
Three colored balls (blue, green, red) move from box A to B in fixed order, and the task requires calculating the probability they return to A in the reverse order RED-GREEN-BLUE, rounded to two decimals. This classic probability puzzle models random draws without replacement from box B back to A. The correct answer is 0.17.
Problem Breakdown
Balls transfer A → B as BLUE → GREEN → BLUE-GREEN → BLUE-GREEN-RED, so box B holds exactly these three distinct balls. Reverse transfer means drawing from B successively without replacement: first RED, then GREEN, then BLUE back to A.
Total possible draw sequences from B: since 3 distinct balls drawn without replacement, there are 3! = 6 equally likely permutations. Only one matches the reverse order (RED, GREEN, BLUE).
Thus, probability = 1/6 ≈ 0.1667, which rounds to 0.17.
Step-by-Step Calculation
Consider sequential probabilities without replacement:
- Probability first draw is RED: 1/3
- Given RED drawn, probability second is GREEN: 1/2
- Given first two, probability third is BLUE: 1/1
Overall: P = 1/3 × 1/2 × 1 = 1/6 ≈ 0.17
All Possible Outcomes
The 6 equiprobable sequences from B, with favorable (reverse) marked:
| Sequence | Order Matches Reverse? |
|---|---|
| R-G-B | Yes |
| R-B-G | No |
| G-R-B | No |
| G-B-R | No |
| B-R-G | No |
| B-G-R | No |
Only 1 out of 6 succeeds, confirming 1/6 = 0.17.
Common Misconceptions
Some mistake this for with-replacement (probability (1/3)³ = 0.04), but successive transfers without replacement dictate permutations. Others overlook distinct colors, but problem specifies unique blue, green, red. No options provided, but answer fills numerical blank as 0.17.
