Question:
Three balls colored blue, green and red are transferred from box A to box B in the order BLUE–GREEN–RED. The probability that they are transferred back from box B to box A in the same order (rounded off to two decimal places) is \(\underline{\hspace{1cm}}\).
Three balls colored blue, green and red are transferred from box A to box B in the order BLUE–GREEN–RED. The probability that they are transferred back from box B to box A in the same order (rounded off to two decimal places) is \(\underline{\hspace{1cm}}\).
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For order-based problems, use permutations: only 1 arrangement matches an exact sequence.
Updated On: Jan 2, 2026
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Correct Answer: 0.16
Solution and Explanation
There are 3 balls, so the total number of possible permutations of drawing them is:
\[ 3! = 6 \]
Only one of these corresponds to the exact order BLUE–GREEN–RED.
Thus, the required probability is:
\[ P = \frac{1}{6} \approx 0.1667 \approx 0.17 \]
\[ 3! = 6 \]
Only one of these corresponds to the exact order BLUE–GREEN–RED.
Thus, the required probability is:
\[ P = \frac{1}{6} \approx 0.1667 \approx 0.17 \]
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