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
(In reply to
re: Solution by Charlie)
If Bernardo picks the lowest number of his 8's set, he is greater than none of Sylvia's numbers. The odds ore 0/n he is greater.
If he picked the second lowest member, he is greater than one of Sylvia's numbers. The odds ore 1/n he is greater.
All the way until he picks the highest member of the set, in which case he is greater than n-1 of Sylvia's numbers. The odds are (n-1)/n he is greater.
Each of those possibilities happens 1/n of the time on average. The total probability is sum (1/n)(P[n sub i]). This comes out to (1/n)(n)(n-1)/2n = (n-1)/2n = 1/2 - 1/2n
The odds are slightly less than half by 1/2n. = 1/2(56)= 1/112
P=55/112 for the 8's part
total P=28/84(1)+56/84(55/112)=28/84(112/112)+56/84(55/112) = (28*112+56*55)/(84*112) = 6216/
9408=
0.66
Edited on July 29, 2015, 2:22 pm
Edited on July 29, 2015, 2:23 pm
Edited on July 29, 2015, 2:27 pm
re(2): Solution
