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29 Mar 2019, 04:02
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (03:04) correct
74%
(02:43)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
Bernardo randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?
(A) 47/72
(B) 37/56
(C) 2/3
(D) 49/72
(E) 39/56
(A) 47/72
(B) 37/56
(C) 2/3
(D) 49/72
(E) 39/56
29 Mar 2019, 05:39
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Bunuel
B has a chance of picking 9 and then he would always have a greater integer. the digit 9 could be picked in any of the 3 picks, so probability of picking 9 is \(\frac{3}{9}=\frac{1}{3}\).
Now probability of not picking 9 is 2/3.
So, when we pick any of three from 1 to 8, there is a probability of two cases
A) the number picked up by the two to be same
ways to pick numbers 8C3 = 8*7*6/(3*2)=56. And one way in these 56, so probability 1/56.
B) the number picked up by B is greater
Now, for same, the probability is 1/56, so the probability of any one is greater is 55/56. Half of these will have B greater and half S greater. Thus P of B greater = \(\frac{1}{2}*\frac{55}{56}=\frac{55}{112}\).
Total probability - \((\frac{1}{3})+(\frac{2}{3}*\frac{55}{112})=\frac{1}{3}+\frac{55}{168}=\frac{37}{56}\)
B
General Discussion
09 Apr 2026, 09:15
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