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14 Oct 2019, 09:06
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:43) correct
60%
(02:21)
wrong
based on 72
sessions
History
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Sixteen persons - \(P_1\) to \(P_{16}\)- are to be seated at a square table, which has four chairs along each side. What is the probability that \(P_7\) and \(P_{13}\) sit on two adjacent chairs on the same side?
A. 1/4
B. 1/10
C. 2/15
D. 1/20
E. 1/25
A. 1/4
B. 1/10
C. 2/15
D. 1/20
E. 1/25
14 Oct 2019, 18:34
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nick1816
On each side, there are 4 chairs, in which there are 3 ways such that P7 and P13 sit adjacent...1-2, 2-3, 3-4
Since there are 4 such sides of the square table, we have 3*4 or 12 ways. But since order matters, the ways = 12*2=24
Total ways to arrange P7 and P13 in any two chairs = 16*15=240
Probability = \(\frac{24}{16*15}=\frac{24}{240}=\frac{1}{10}\)
B
General Discussion
14 Oct 2019, 21:35
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nick1816
Using combination:
*Four people are seated in one side, there are 3C2 ways or 3 ways to select 2 person who always sit together.
Four side are there so total 3*4-12 ways
*Total way for selecting 2 people 16C2- 120 ways
Probability = 12/120 = 1/10 way
Can anyone solve this using Probability Method?
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