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From the 5 points A, B, C, D, and E on the number line above

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From the 5 points A, B, C, D, and E on the number line above [#permalink] New post 23 Jun 2016, 03:23
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From the 5 points A, B, C, D, and E on the number line above, 3 different points are to be randomly selected. What is the probability that the coordinates of the 3 points selected will all be positive?


A. \(\frac{1}{10}\)

B. \(\frac{1}{5}\)

C. \(\frac{3}{10}\)

D. \(\frac{2}{5}\)

E. \(\frac{3}{5}\)


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Question: 9
Page: 120
[Reveal] Spoiler: OA

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Re: From the 5 points A, B, C, D, and E on the number line above [#permalink] New post 24 Jun 2016, 10:59
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Explanation

Start with a translation and work from there:

P(all 3 numbers are positive) = P(1st # is positive AND 2nd # is positive AND 3rd # is positive)
= P(1st # is positive) x P(2nd # is positive) x P(3rd # is positive)
= 3/5 x 2/4 x 1/3
= 1/10
=
[Reveal] Spoiler:
A



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Re: From the 5 points A, B, C, D, and E on the number line above [#permalink] New post 27 Jun 2016, 14:27
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p(1st posi point)=1/5
p(2nd;;;)=1/4
p(3rd;;;;;)=1/3
p(3 posit point)=1/5*1/4*1/3*6
Ans:1/10
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Re: From the 5 points A, B, C, D, and E on the number line above [#permalink] New post 27 Jun 2016, 17:55
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There can be only one positive coordinate of three point, and there are 5C3 ways of selecting three points out of 5 points. Hence the probability is 1/5C3 = 1/10
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Re: From the 5 points A, B, C, D, and E on the number line above [#permalink] New post 29 Jun 2020, 17:54
All possibilities are (5)(4)(3) = 60, so there are 60 possibilities, but there are only 6 ways to obtain all positive (CDE,CED,DEC,DCE,ECD,EDC). So the probability to obtain all positive is 6/60 equals to 1/10.
Re: From the 5 points A, B, C, D, and E on the number line above   [#permalink] 29 Jun 2020, 17:54
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From the 5 points A, B, C, D, and E on the number line above

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