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jaskarannagra
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AMS employs 8 professors on their staff. Their respective probability Updated on: 09 Mar 2015, 02:35
Originally posted by jaskarannagra on 07 May 2013, 04:05.
Last edited by Bunuel on 09 Mar 2015, 02:35, edited 1 time in total.
Renamed the topic and edited the question.
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95% (hard)

Question Stats:

36% (02:28) correct 64% (02:05) wrong based on 80 sessions

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AMS employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. The probability that after 10 years at least 6 of them still work in AMS is ?

A. 0.19
B. 1.22
C. 0.1
D. 0.16
E. none of these
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A
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KarishmaB
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AMS employs 8 professors on their staff. Their respective probability 07 May 2013, 10:37
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jaskarannagra
Ams employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years atleast 6 of them still work in Ams is ???

a) 0.19 b) 1.22 c) 0.1 d) 0.16
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Responding to a pm:

Not a GMAT question. Too many calculations.

1 = P(No one remains) + P(1 remains) + P(2 remain) + ..... P(6 remain) + P(7 remain) + P(All 8 remain)

P(At least 6 are still there) = P(6 remain) + P(7 remain) + P(8 remain)

P(6 remain) = .2*.3*.4*.5*.6*.7*.2*.1 + .2*.3*.4*.5*.6*.3*.2*.9 + .....
P(7 remain) = ...
P(All 8 remain) = .2*.3*.4*.5*.6*.7*.8*.9

Way too many calculations!
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Narenn
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AMS employs 8 professors on their staff. Their respective probability 07 May 2013, 12:30
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jaskarannagra
Ams employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years atleast 6 of them still work in Ams is ???

a) 0.19 b) 1.22 c) 0.1 d) 0.16
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Yeah! As karishma wrote, This can never be a GMAT (or any other time constrained test) question

Probability = 6 remains + 7 remains + 8 remains

6 Remains = 8C6 (Creation of 28 different pairs of 6 peoples X multiplication of their respective probabilities 28 times)

7 Remains = 8C7 (Creation of 8 different pairs of 7 peoples X multiplication of their respective probabilities 8 times)

8 Remains = 8C8 (0.2 * 0.3 * 0.4 * 0.5 * 0.6 * 0.7 * 0.8 * 0.9)

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jaskarannagra
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AMS employs 8 professors on their staff. Their respective probability 08 May 2013, 01:01
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jaskarannagra
Ams employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years atleast 6 of them still work in Ams is ???

a) 0.19 b) 1.22 c) 0.1 d) 0.16

Yeah! As karishma wrote, This can never be a GMAT (or any other time constrained test) question

Probability = 6 remains + 7 remains + 8 remains

6 Remains = 8C6 (Creation of 28 different pairs of 6 peoples X multiplication of their respective probabilities 28 times)

7 Remains = 8C7 (Creation of 8 different pairs of 7 peoples X multiplication of their respective probabilities 8 times)

8 Remains = 8C8 (0.2 * 0.3 * 0.4 * 0.5 * 0.6 * 0.7 * 0.8 * 0.9)

Narenn
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Sure! Its not a GMAT test question.....But "or any other time constrained test" is falsely assumed... It is a question from CAT , which is an mba entrance test in india... Anyways.. I posted on this forum for the sake of learning and also because this forum has fastest replies of all... Kudos for that....

So, My question is that is there any faster way of solving such a question with multiple different probabilities.... can we take average of probabilities in any case whatsoever?
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AMS employs 8 professors on their staff. Their respective probability 09 Mar 2015, 00:56
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here 8 professors are there.
Asking atleast 6 of them continue ,
it has 3 cases.
1 all 8 professors continue.
2 7 professors continue and 1 professors discontinue.
3 6 professors continue and 2 professors discontinue.

1st case all 8 continue is = 2/10*3/10*4/10*5/10*6/10*7/10*/8/10*9/10=9factorial/10power8
=362880/100000000
=>0.00363.

2nd case any 7 professors continue, and 1out of 8 discontinue ,8C1 means 8 ways.
= 2/10*3/10.......8/10*1/10, (9/10 prodbability professor discontinue then 1/10)
in this way if we calculate for 8 possibilities then value is =>0.03733.


3rd case any 6 professors continue and 2out of 8 discontinue , 8C2 means 28 ways.

= 2/10*3/10.....2/10*1/10( 2 professors discontinue.

if we calculate for 28 possibilites P value is=>0.1436

=0.00363+0.03733+0.1436=0.18456=
0.19
Option A sufficient.