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Mr. A randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and arranges them in decreasing order to from a 3 – digit number. Mr. B picks randomly 3 distinct digits from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in decreasing order to form a 3-digit number. Then which is/are correct

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The probability that Mr. A’s number is larger than Mr. B’s number is $\frac{37}{56}$

The probability that Mr. A’s number is larger than Mr. B’s number is $\frac{9}{14}$

The probability that A and B has the same 3-digit number is $\frac{1}{84}$

The probability that A and B has the same 3-digit number is $\frac{1}{56}$

answer is A, C.

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Detailed Solution

$A) P (E) = \frac{(^{1} C_{1} \times^{8} C_{2})^{8} C_{3} + (^{8} C_{3}) C_{2}}{^{9} C_{2} \times^{8} C_{3}} = \frac{37}{56} C) P (E) = \frac{^{8} C_{3}}{^{9} C_{2} \times^{8} C_{3}} = \frac{1}{84}$

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