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# A number is randomly chosen from a list of 10 consecutive positive int

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Math Expert
Joined: 02 Sep 2009
Posts: 49968
A number is randomly chosen from a list of 10 consecutive positive int  [#permalink]

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31 Jul 2018, 00:12
00:00

Difficulty:

15% (low)

Question Stats:

81% (00:51) correct 19% (00:56) wrong based on 53 sessions

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A number is randomly chosen from a list of 10 consecutive positive integers. What is the probability that the number selected is greater than the average (arithmetic mean) of all 10 integers?

A. $$\frac{3}{10}$$

B. $$\frac{2}{5}$$

C. $$\frac{1}{2}$$

D. $$\frac{7}{10}$$

E. $$\frac{4}{5}$$

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Re: A number is randomly chosen from a list of 10 consecutive positive int  [#permalink]

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31 Jul 2018, 00:52
Bunuel wrote:
A number is randomly chosen from a list of 10 consecutive positive integers. What is the probability that the number selected is greater than the average (arithmetic mean) of all 10 integers?

A. $$\frac{3}{10}$$

B. $$\frac{2}{5}$$

C. $$\frac{1}{2}$$

D. $$\frac{7}{10}$$

E. $$\frac{4}{5}$$

average of n consecutive int = (1st term + nth term)/2
say 1,2,3,4,5
avg=6/2=3
take any such sequence say 1...10
avg=11/2=5.5
favourable outcomes = 6,7,8,9,10 = 5
total no of outcomes = 10
probability = 5/10=1/2
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Re: A number is randomly chosen from a list of 10 consecutive positive int  [#permalink]

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31 Jul 2018, 03:00
Bunuel wrote:
A number is randomly chosen from a list of 10 consecutive positive integers. What is the probability that the number selected is greater than the average (arithmetic mean) of all 10 integers?

Now,
Average of 10 consecutive positive integers = Median of 10 consecutive positive integers

Median of 10 consecutive positive integers = (5th term + 6th term)/ 2
The result will always lower than half of the elements and greater than remaining other half of the elements
Therefore probability = 1/2

Hence, C.
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Re: A number is randomly chosen from a list of 10 consecutive positive int &nbs [#permalink] 31 Jul 2018, 03:00
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