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.
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