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?
Source: AMO
The numbers are arranged in descending order.
For Bernardo, there are there are 8*7/2 numbers starting with a 9, since he has a choice of 8 numbers for one number, 7 numbers for the other number, and they have to be arranged so as to be in descending order. 8*7/2=28. Every one of these numbers is guaranteed to be greater than any number in Sylvia's set.
The rest of Bernardo's set is exactly the same as Sylvia's. There is a 50/50 chance that Bernardo has the larger number when drawing from this portion of his set.
The total size of Sylvia's set is
(7*6+6*5+5*4+4*3+3*2+2*1)/2 = 56 numbers.
Bernardo has 56+28 = 84 numbers.
He draws from the 9's portion of his set 28/84 of the time with probability 1 that he has the larger number. He draws from the rest of his set 56/84 of the time with a 0.5 probability of having the larger number. The total probability is:
28/84(1)+56/84(0.5) = 56/84 = 2/3 of the time.
Solution
