Problem Solving

If you select two cards from a pile of cards numbered 1 to 10, what is the probability that the sum of the numbers is less than the average of the pile?

(A) 1/100

(B) 2/45

(C) 2/25

(D) 4/45

(E) 1/10

OA: D

P.S: I did it by considering total outcomes and actual outcomes. Would like to know how it can be solved in other way - I mean, taking 1st card(and counting probabilities) and 2nd card(and counting probabilities) while considering the constraint that sum must be less than average.

RBBmba@2014 wrote:If you select two cards from a pile of cards numbered 1 to 10, what is the probability that the sum of the numbers is less than the average of the pile?

(A) 1/100

(B) 2/45

(C) 2/25

(D) 4/45

(E) 1/10

The 10 cards -- which are composed of the consecutive integers 1 through 10, inclusive -- constitute an EVENLY SPACED SET.
For any evenly spaced set, average = (biggest + smallest)/2.
Thus, the average value of the 10 cards = (10+1)/2 = 5.5.
Implication:
A good outcome occurs when the sum of the 2 cards is 5 or less.

Approach 1:
P = good/all.

All possible outcomes:
Number of options for the 1st card = 10.
Number of options for the 2nd card = 9.
To combine these options, we multiply:
10*9 = 90.

Good outcomes:
1, 2
1, 3
1, 4
2, 1
2, 3
3, 1
3, 2
4, 1.
Total good outcomes = 8.

Thus:
P = 8/90 = 4/45.

The correct answer is D.

Approach 2:
P(1st card is 1 and 2nd card is 2, 3, or 4) = 1/10 * 3/9 = 3/90.
P(1st card is 2 and 2nd card is 1 or 3) = 1/10 * 2/9 = 2/90.
P(1st card is 3 and 2nd card is 1 or 2) = 1/10 * 2/9 = 2/90.
P(1st card is 4 and 2nd card is 1) = 1/10 * 1/9 = 1/90.
Since any of the options above would constitute a good outcome, we ADD the fractions:
3/90 + 2/90 + 2/90 + 1/90 = 8/90 = 4/45.

The correct answer is D.

Why does the order matter?

It doesn't make sense to answer after the OA is given. But still, my guess is D.

RBBmba@2014 wrote: All possible outcomes:
Number of options for the 1st card = 10.
Number of options for the 2nd card = 9.
To combine these options, we multiply:
10*9 = 90.

Can you please explain this part ?

talaangoshtari wrote:Why does the order matter?

In my first solution, I calculated the following:
(good orderings)/(all possible orderings).

It is also possible to disregard order and count the following:
(good combinations)/(all possible combinations).

Good combinations:
1, 2
1, 3
1, 4
2, 3.
Total options = 4.

All possible combinations:
From 10 cards, the number of ways to choose a combination of 2 = 10C2 = (10*9)/(2*1) = 45.

Thus:
(good combinations)/(all possible combinations) = 4/45.

sanchay wrote:

All possible outcomes:
Number of options for the 1st card = 10.
Number of options for the 2nd card = 9.
To combine these options, we multiply:
10*9 = 90.

Can you please explain this part ?

FUNDAMENTAL COUNTING PRINCIPLE:
If one event has m possible outcomes, and a second, independent event has n possible outcomes, then there are m*n total possible ways for the events to happen together.

Problem above:
Since there are 10 possible outcomes for the first card, and 9 possible outcomes for the second card, there are 10*9 = 90 ways that the first two cards can be chosen.