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In a 10 member club including M, there are 3 posts of: President, Secr
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08 Mar 2021, 02:25
08 Mar 2021, 02:25
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Question Stats:
25% (00:44) correct
75% (02:03) wrong based on 4 sessions
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In a 10 member club including M, there are 3 posts of: President, Secretary, and Treasurer. A President is to elected from these 10 members, A Secretary from 9 and a Treasurer from 8 members. What is the Probability of M becoming;
I. President only
II. Secretary only
III. Treasurer only
I. President only
II. Secretary only
III. Treasurer only
Show: ::
I. \(\frac{7}{90}\) II. \(\frac{7}{80}\) III. \(\frac{1}{10}\)
Re: In a 10 member club including M, there are 3 posts of: President, Secr
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08 Mar 2021, 03:06
08 Mar 2021, 03:06
1
Probability = \(\frac{Favorable \space Outcomes}{Total \space Outcomes}\)
(I) President only mean: P(President) * P(Not Secretary) * P(Not Treasurer)
P(President) = \(\frac{1}{10}\)
Now since he is out of the race for the president, there are 9 remaining members. The probability that he would have been made Secretary also would be \(\frac{1}{9}\).
Therefore P(Not Secretary) = \(1 - \frac{1}{9} = \frac{8}{9}\)
Similarly, P(treasurer) = \(\frac{1}{8}\) and P(not treasurer) = \(1 - \frac{1}{8} = \frac{7}{8}\)
Therefore P(President Only) = \(\frac{1}{10} * \frac{8}{9} * \frac{7}{8} = \frac{7}{90}\)
(II) Secretary only means: P(Not President) * P(Secretary) * P(Not Treasurer)
P(President) = \(\frac{1}{10}\). Therefore P(Not President) = \(1 - \frac{1}{10} = \frac{9}{10}\)
P(Secretary) = \(\frac{1}{9}\).
P(Treasurer) = \(\frac{1}{8}\) and P(Not Treasurer) = \(1 - \frac{1}{8} = \frac{7}{8}\)
Therefore P(Secretary Only) = \(\frac{9}{10} * \frac{1}{9} * \frac{7}{8} = \frac{7}{80}\)
(III) Treasurer only means: P(Not President) * P(Not Secretary) * P(Treasurer)
P(President) = \(\frac{1}{10}\). Therefore P(Not President) = \(1 - \frac{1}{10} = \frac{9}{10}\)
P(Secretary) = \(\frac{1}{9}\). P(Not Secretary) = \(1 - \frac{1}{9} = \frac{8}{9}\)
P(Treasurer) = \(\frac{1}{8}\)
Therefore P(Treasurer Only) = \(\frac{9}{10} * \frac{8}{9} * \frac{1}{8} = \frac{1}{10}\)
Arun Kumar
(I) President only mean: P(President) * P(Not Secretary) * P(Not Treasurer)
P(President) = \(\frac{1}{10}\)
Now since he is out of the race for the president, there are 9 remaining members. The probability that he would have been made Secretary also would be \(\frac{1}{9}\).
Therefore P(Not Secretary) = \(1 - \frac{1}{9} = \frac{8}{9}\)
Similarly, P(treasurer) = \(\frac{1}{8}\) and P(not treasurer) = \(1 - \frac{1}{8} = \frac{7}{8}\)
Therefore P(President Only) = \(\frac{1}{10} * \frac{8}{9} * \frac{7}{8} = \frac{7}{90}\)
(II) Secretary only means: P(Not President) * P(Secretary) * P(Not Treasurer)
P(President) = \(\frac{1}{10}\). Therefore P(Not President) = \(1 - \frac{1}{10} = \frac{9}{10}\)
P(Secretary) = \(\frac{1}{9}\).
P(Treasurer) = \(\frac{1}{8}\) and P(Not Treasurer) = \(1 - \frac{1}{8} = \frac{7}{8}\)
Therefore P(Secretary Only) = \(\frac{9}{10} * \frac{1}{9} * \frac{7}{8} = \frac{7}{80}\)
(III) Treasurer only means: P(Not President) * P(Not Secretary) * P(Treasurer)
P(President) = \(\frac{1}{10}\). Therefore P(Not President) = \(1 - \frac{1}{10} = \frac{9}{10}\)
P(Secretary) = \(\frac{1}{9}\). P(Not Secretary) = \(1 - \frac{1}{9} = \frac{8}{9}\)
P(Treasurer) = \(\frac{1}{8}\)
Therefore P(Treasurer Only) = \(\frac{9}{10} * \frac{8}{9} * \frac{1}{8} = \frac{1}{10}\)
Arun Kumar
Re: In a 10 member club including M, there are 3 posts of: President, Secr
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11 Mar 2021, 12:11
11 Mar 2021, 12:11
are you sure these are gre questions,from any angle don't feel like it!!!
its more like CAT/GMAT type
its more like CAT/GMAT type
