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P, Q, and R each try to execute a job and create a report on it. The
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Updated on: 12 Aug 2018, 23:02
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eGMAT Question of the Week #6P, Q, and R each try to execute a job and create a report on it. The individual probabilities for their completion of the jobs are \(\frac{1}{3}\), \(\frac{2}{3}\), and \(\frac{3}{5}\) and the probability for any of them not to finish the report is \(\frac{2}{5}\). If one can write the report only after finishing the job, then what is the probability that only P and Q will complete their jobs and finish their reports? A. \(\frac{4}{625}\)
B. \(\frac{4}{125}\)
C. \(\frac{32}{625}\)
D. \(\frac{4}{25}\)
E. \(\frac{32}{125}\) To access all the questions: Question of the Week: Consolidated List
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Re: P, Q, and R each try to execute a job and create a report on it. The
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Updated on: 09 Jul 2018, 19:39
P, Q, and R each try to execute a job and create a report on it. Completing the job and finishing the report by P, Q and R are independent of each other. GIven probability for any of them not finishing the report is \(\frac{2}{5}\) probability for any of them finishing the report is \(1\frac{2}{5} = \frac{3}{5}\) To find out the probability that only P and Q will complete their jobs and finish their reports We need to find Required probability is the product of three of the following probabilitites P(P will complete his job AND finish the report) P(Q will complete his job AND finish the report) P(R will not complete his job) OR P(R will complete his job AND not finish the report) probability that P will complete his job and finish the report = \(\frac{1}{3} * \frac{3}{5} = \frac{1}{5}\) probability that Q will complete his job and finish the report =\(\frac{2}{3} * \frac{3}{5} = \frac{2}{5}\) probability that R will not complete his job = \(1  \frac{3}{5} = \frac{2}{5}\) probability that R will complete his job and not finish the report = \(\frac{3}{5} * \frac{2}{5} = \frac{6}{25}\) Substituting the above values, we get \((\frac{1}{5})(\frac{2}{5})(\frac{2}{5} + \frac{6}{25}) = \frac{32}{625}\) Hence option C
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Originally posted by workout on 06 Jul 2018, 20:40.
Last edited by workout on 09 Jul 2018, 19:39, edited 1 time in total.




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P, Q, and R each try to execute a job and create a report on it. The
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Updated on: 12 Aug 2018, 23:35
Hey everyone, We will post the solution very soon. Till then, try it one more time and post your analysis.
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Re: P, Q, and R each try to execute a job and create a report on it. The
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Updated on: 09 Jul 2018, 10:55
Given that:
Probability that \(P\) execute the job = \(\frac{1}{3}\) Probability that \(Q\) execute the job = \(\frac{2}{3}\) Probability that \(R\) execute the job = \(\frac{3}{5}\)
Probability that anyone not finishing the report = \(\frac{2}{5}\) => Probability that anyone finish the report = 1  \(\frac{2}{5}\) = \(\frac{3}{5}\)
Given that only \(P\) and \(Q\) will execute the job and finish the report. So,\(R\) will not execute the job
Therefore Probability of \(R\) not executing the job = 1 \(\frac{3}{5}\) = \(\frac{2}{5}\)
Required probability = (P execute the job and finish report) x ( Q execute the job and finish the report) x (R not executing the job) = \((\frac{1}{3} * \frac{3}{5}) ( \frac{2}{3} * \frac{3}{5}) (\frac{2}{5})\) = \(\frac{1}{5} * \frac{2}{5} * \frac{2}{5}\) = \(\frac{4}{125}\) Answer = B



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Re: P, Q, and R each try to execute a job and create a report on it. The
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09 Jul 2018, 09:12
I did go with C but the OA says B. So here is why it could be B.
P(E) for P and Q to complete their jobs and reports = [(P(A)*P(B)*P(C) all completing the Job) * (P(A)*P(B) completing the Report *P(C) not completing the Report)] + [(P(A)*P(B) completing the Job * P(C) not completing the Job ) * (P(A)*P(B) completing the Report *P(C) not completing the Report)].
We should remember that when C does not complete this job it also implies that he cannot complete his report.
Substituting the values we get >[\((1/3*2/3*3/5)* (3/5*3/5*2/5) = 12/625\)] + [\((1/3*2/3*2/5)* (3/5*3/5*2/5) = 8/625\)] = \(20/625\) = \(4/125\)
Do let me know if my understanding is wrong.
Aditya. Do give kudos if you find this answer helpful.



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Re: P, Q, and R each try to execute a job and create a report on it. The
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09 Jul 2018, 10:42
The reason why I calculated B and not C is the last part of the question: then what is the probability that only P and Q will complete their jobs and finish their report For me completing the job and finishing their report implies that R was not able to complete the job. Therefore, it is not necessary to take into account, that R could have completed the job but not the report.



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P, Q, and R each try to execute a job and create a report on it. The
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Updated on: 12 Aug 2018, 23:33
Solution Given:• P, Q, and R each try to execute a job and create a report on it • The individual probabilities for their completion of the jobs are \(\frac{1}{3}\), \(\frac{2}{3}\), and \(\frac{3}{5}\) • The probability for any of them not to finish the report is \(\frac{2}{5}\)
o Hence, the probability of finishing the report = \((1 – \frac{2}{5}) = \frac{3}{5}\) • One can write the report only after finishing the job To find:• The probability that only P and Q will complete their jobs and finish their reports Approach and Working: it is given that, as per the favourable event, only P and Q will complete their jobs and finish their reports • The probability that P will execute the job and finish the report = \(\frac{1}{3} * \frac{3}{5} = \frac{1}{5}\) • The probability that Q will execute the job and finish the report = \(\frac{2}{3} * \frac{3}{5} = \frac{2}{5}\) Now, as only P and Q will complete their jobs and finish their reports, it also means, there are two possibilities exist for R • Either R will not finish the job, and definitely not finish the report (as one cannot finish the report without executing the job)
o Probability of this event = \((1 – \frac{3}{5}) * \frac{2}{5} = \frac{2}{5} * \frac{2}{5} = \frac{4}{25}\) • Or else, R will execute the job but will not finish the report
o Probability of this event = \(\frac{3}{5} * \frac{2}{5} = \frac{6}{25}\) So, we can say, • Probability (only P and Q will complete their jobs and finish their reports) = P (P completes job & finish report) AND P (Q completes job & finish report) AND [P (R does not complete the job & does not complete the report) OR P (R does complete the job & does not complete the report) = \(\frac{1}{5} * \frac{2}{5} * [\frac{4}{25} + \frac{6}{25}]\) = \(\frac{1}{5} * \frac{2}{5} * \frac{2}{5}\) = \(\frac{4}{125}\) Hence, the correct answer is option B. Answer: B
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Re: P, Q, and R each try to execute a job and create a report on it. The
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21 Jul 2018, 09:56
Hi expert,
I can not understand why we need to consider the probability of R as the question asks to find the probability that only P and Q will complete their jobs and finish their reports.
Plz help.



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Re: P, Q, and R each try to execute a job and create a report on it. The
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21 Jul 2018, 14:18
Hello expert, request you explain this how do we consider the probability Sent from my CPH1727 using GMAT Club Forum mobile app



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Re: P, Q, and R each try to execute a job and create a report on it. The
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22 Jul 2018, 05:38
BARUAH wrote: Hi expert,
I can not understand why we need to consider the probability of R as the question asks to find the probability that only P and Q will complete their jobs and finish their reports.
Plz help. Hey BARUAH, If you read the question carefully, the favourable event is defined as " only P and Q will complete their jobs and finish their reports." It means we have to ensure that R cannot complete the job and finish the report. For this reason we have to consider the probability of R. Hope this answers your query.
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Re: P, Q, and R each try to execute a job and create a report on it. The
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22 Jul 2018, 05:40
Anshup wrote: Hello expert, request you explain this how do we consider the probability Sent from my CPH1727 using GMAT Club Forum mobile appHey Anshup, Can you provide a little bit more details about you query? The answer we provided takes all the probability values given in the question only.
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Re: P, Q, and R each try to execute a job and create a report on it. The
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26 Jul 2018, 21:09
Hello EgmatQuantExpert, thank you for the detailed explanation. My mistake was to miss the probability of not finishing the report when R did not even finish his job. I thought that since we are told that one cannot even start his report if he didn't finish his job, hence there is no need to multiply 2/5 (13/5  probability that R will not finish his job) with 2/5 (probability of not finishing a report). Please help me understand why we need to take into consideration the probability of not finishing a report when we know that R did jot finish his job, hence he cannot even start his report? Thanks a lot!



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Re: P, Q, and R each try to execute a job and create a report on it. The
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07 Mar 2019, 21:07
Hello BunuelCould you please look at this question. It seems that answer should be C.



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Re: P, Q, and R each try to execute a job and create a report on it. The
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09 Apr 2019, 05:02
EgmatQuantExpert wrote: Solution Now, as only P and Q will complete their jobs and finish their reports, it also means, there are two possibilities exist for R • Either R will not finish the job, and definitely not finish the report (as one cannot finish the report without executing the job)
o Probability of this event = \((1 – \frac{3}{5}) * \frac{2}{5} = \frac{2}{5} * \frac{2}{5} = \frac{4}{25}\) Answer: B to be honest , I don't understand why we have to multibly (r not finishing the job ) * (r not finishing the report ) in the first case ,since if he didn't finish the job he can't even write the report , so the event will stop there by multiblying we make the probability smaller than its actual value



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Re: P, Q, and R each try to execute a job and create a report on it. The
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28 May 2020, 05:43
foryearss wrote: EgmatQuantExpert wrote: Solution Now, as only P and Q will complete their jobs and finish their reports, it also means, there are two possibilities exist for R • Either R will not finish the job, and definitely not finish the report (as one cannot finish the report without executing the job)
o Probability of this event = \((1 – \frac{3}{5}) * \frac{2}{5} = \frac{2}{5} * \frac{2}{5} = \frac{4}{25}\) Answer: B to be honest , I don't understand why we have to multibly (r not finishing the job ) * (r not finishing the report ) in the first case ,since if he didn't finish the job he can't even write the report , so the event will stop there by multiblying we make the probability smaller than its actual value Agree. Can any expert shed more light on this?



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P, Q, and R each try to execute a job and create a report on it. The
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28 May 2020, 08:40
EgmatQuantExpertand in addition, why i think the ans B is impossible. If u draw a probability tree for R, (3/5)(2/5)+(3/5)(3/5)+(2/5)(2/5) does not equal to 1. Since u said its impossible for R to do report if job is not finished, so (3/5)(2/5)+(3/5)(3/5)+(2/5)(2/5)+ (2/5)(3/5)=1 is not possible. so the probability tree should be: (3/5)(2/5)+(3/5)(3/5)+(2/5) =1. can someone please clarify if my understanding is correct?
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