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A certain box has 10 cards written integers from 1 to 10

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by AAPL » Wed Apr 25, 2018 5:54 am

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Answers

A

B

C

D

E

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Difficulty—

A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of the numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15

The OA is C.

Average of 10 cards = (1 + 10)/2 = 5.5

Number of ways of picking 2 cards from 10 cards = 10C2 = 45

Possible ways of picking 2 cards such that sum will be less than 5.5 = (1 + 2), (1 + 3), (1 + 4), (2 + 3) = 4

Probability = 4/45

Has anyone another strategic approach to solve this PS question? Thanks!
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by Brent@GMATPrepNow » Wed Apr 25, 2018 6:38 am
AAPL wrote:A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of the numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15
As AAPL has shown, the average value of the ten cards is 5.5
So, we want to find P(sum of 2 cards is less than 5.5)

We can use combinations to determine the umber of ways to select 2 of the 10 cards
We can select 2 cards from 10 cards is 10C2 ways (= 45 ways)

When we examine the answer choices, and view them all with denominator 45, we get: 2/45, 3/45, 4/45, 5/45and 6/45
This tells us that the NUMBER of ways to get a sum that's less than 5.5 must range from 2 to 6.
Given such small numbers, it makes sense to just list the possible outcomes that yield a sum that's less than 5.5.

The possible outcomes are:
1 & 2
1 & 3
1 & 4
2 & 3
DONE!

So, there are only 4 possible outcomes that yield a sum that's less than 5.5.

P(sum of 2 cards is less than 5.5) = 4/45

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Wed Apr 25, 2018 10:07 am
Average in total = (1+2+...+9+10) / 10 = 5.5. The smaller pair than 5.5 is (1,2), (1,3), (1,4), (2,0), which makes 4 cases. Thus, probability = 4C1/10C2 = 4/45. Option C.

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by Scott@TargetTestPrep » Thu Apr 26, 2018 2:47 pm
AAPL wrote:A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of the numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15
The number of ways to select 2 cards from 10 is 10C2 = 10!/[2! x (10 - 2)!] = 10!/(2! x 8!) = (10 x 9)/2! = 90/2 = 45.

The average of the integers from 1 to 10, inclusive, is calculated by using the shortcut formula for the average of an evenly-spaced set: average = (smallest value + largest value)/2 = (1 + 10)/2 = 5.5.

The following are all possible pairs of two numbers whose sum is less than 5.5:

{1, 2}, {1, 3}, {1, 4}, and {2, 3}

Thus, the probability that the sum of two numbers drawn is less than the average of the 10 numbers on the cards is 4/45.

Answer: C

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[email protected]

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