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

Question

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 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

* A solution will be posted in two days.

Kurtosis · Answer

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

Answer: C

MathRevolution · Answer

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 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

* A solution will be posted in two days.

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

-> Average in total=(1+2+…+9+10)/10=5.5. The smaller pairs than 5.5 is (1,2),(1,3),(1,4),(2,0), which makes 4 cases. Thus, probability=4C1/10C2=4/45. The answer is C.

TheMechanic · Answer

Why didn't we consider the order of picking the cards. Say we select (1,4) or (4,1): that constitutes two combinations. Although I know the numerator and denominator would cancel each other out but shouldn't the actual fraction be : 8/90? Please advise.

Thanks.

Thanks.

BrentGMATPrepNow · Answer

As others have shown above, 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/ 45  and 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

ScottTargetTestPrep · Answer

The number of ways to select 2 cards from 10 is 10C2 = 10!/  = 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

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

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

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