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**