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A certain class consists of 8 students, including Kim. Each day, three
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17 Feb 2012, 07:50
17 Feb 2012, 07:50
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A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
Source: Jeff Sackman questions - https://www.gmathacks.com
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
17 Feb 2012, 08:23
17 Feb 2012, 08:23
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Expert Reply
metallicafan wrote:
Bunuel wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
P.S. Please provide answer choices if available.
Thank you bunuel! However, I don't get your approach well. Could you explain your logic? I believe that the probability of picking somenone in the second task is different from picking someone in the first because the number of elements available to pick changes. That's why my approach is step by step.
Please, show me the light!
The wording of the question makes it harder then it is. The question basically asks: what is the probability that Kim is among the first 3 students out of 8, which is 3/8.
One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;
Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
Or: the same with probability approach: \(P=1-\frac{7}{8}*\frac{6}{7}*\frac{5}{6}=\frac{3}{8}\).
As you can see first approach is easiest and shortest one.
P.S. What about the answer choices?
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Re: A certain class consists of 8 students, including Kim. Each day, three
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17 Feb 2012, 07:55
17 Feb 2012, 07:55
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metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
P.S. Please provide answer choices if available.
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
17 Feb 2012, 08:12
17 Feb 2012, 08:12
Bunuel wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
P.S. Please provide answer choices if available.
Thank you bunuel! However, I don't get your approach well. Could you explain your logic? I believe that the probability of picking somenone in the second task is different from picking someone in the first because the number of elements available to pick changes. That's why my approach is step by step.
Please, show me the light!
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
17 Feb 2012, 08:49
17 Feb 2012, 08:49
Thank you Bunuel!, I have updated the post with the choices. 
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
17 Feb 2012, 10:00
17 Feb 2012, 10:00
Thankx Bunuel, the explanation covers all possible ways of solving the problem
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
17 Feb 2012, 17:43
17 Feb 2012, 17:43
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One more way to sove this is to reverse the logic:
NOT chosen on first, second and third pick leads to the following calculation:
7/8 * 6/7 * 5/6 = 0,625. Since we are looking for the probability Kim gets chosen it's 1-0,625 = 0,375 = 3/8
NOT chosen on first, second and third pick leads to the following calculation:
7/8 * 6/7 * 5/6 = 0,625. Since we are looking for the probability Kim gets chosen it's 1-0,625 = 0,375 = 3/8
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
23 Feb 2012, 10:06
23 Feb 2012, 10:06
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I agree with B 3/8
first task = 1/8 chance
second task= 7/8*1/7 = 1/8
third task = 7/8*6/7*1/6 = 1/8
chance that she gets picked for one of these is 1/8+1/8+1/8 or 3/8
first task = 1/8 chance
second task= 7/8*1/7 = 1/8
third task = 7/8*6/7*1/6 = 1/8
chance that she gets picked for one of these is 1/8+1/8+1/8 or 3/8
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
03 May 2014, 23:35
03 May 2014, 23:35
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metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
Hello,
Can someone please explain :
for completion of task B - why 7/8 * 1/7* 6/6??
for completion of task C - why 7/8 * 6/7* 1/6??
plx explain the one in BOLD especially..
Sharmita
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
04 May 2014, 03:01
04 May 2014, 03:01
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Expert Reply
msharmita wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
Hello,
Can someone please explain :
for completion of task B - why 7/8 * 1/7* 6/6??
for completion of task C - why 7/8 * 6/7* 1/6??
plx explain the one in BOLD especially..
Sharmita
Probability of completing Task A:
P(Kim)*P(any)*P(any) = \(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
P(any bu Kim)*P(Kim)*P(any) = \(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
P(any but Kim)*P(any but Kim)*P(any) = \(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Faster solutions are here: a-certain-class-consists-of-8-students-including-kim-127730.html#p1046004 and here: a-certain-class-consists-of-8-students-including-kim-127730.html#p1046020
Similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
new-set-of-mixed-questions-150204-100.html#p1208473
a-medical-researcher-must-choose-one-of-14-patients-to-recei-124775.html
Hope this helps.
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
24 Aug 2016, 08:46
24 Aug 2016, 08:46
Hi Bunuel,
Should the question be read as "What is the probability that Kim will be selected to complete ATLEAST one of the three tasks?"
Or should it be read as "What is the probability that Kim will be selected to complete EXACTLY one of the three tasks?"
I felt the question was slightly ambiguous. If so, will such type of questions appear on the GMAT ?
Should the question be read as "What is the probability that Kim will be selected to complete ATLEAST one of the three tasks?"
Or should it be read as "What is the probability that Kim will be selected to complete EXACTLY one of the three tasks?"
I felt the question was slightly ambiguous. If so, will such type of questions appear on the GMAT ?
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
24 Aug 2016, 08:51
24 Aug 2016, 08:51
Sorry for the confusion. I get it know. Since the left-over tasks are to be done only by the remaining people, one person can only do one task. So in this case both mean the same.
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
12 Dec 2019, 07:40
12 Dec 2019, 07:40
Hi Bunuel ,
Please clarify my below doubts. It will help me a lot in other questions too. I am very much confused.
One can also do: P=C11∗C27C38=38P=C11∗C72C83=38, where C11C11 is # of ways to select Kim, C27C72 is # of ways to select any 2 students out of 7 left and C38C83 is total # of ways to select 3 students from 8;
In your above approach, How you have taken C^1_1 ? I always took C^1_8 as we need to select 1 from 8.
Please reply
Thanks a lot
Please clarify my below doubts. It will help me a lot in other questions too. I am very much confused.
One can also do: P=C11∗C27C38=38P=C11∗C72C83=38, where C11C11 is # of ways to select Kim, C27C72 is # of ways to select any 2 students out of 7 left and C38C83 is total # of ways to select 3 students from 8;
In your above approach, How you have taken C^1_1 ? I always took C^1_8 as we need to select 1 from 8.
Please reply
Thanks a lot
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Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
28 Apr 2020, 09:42
28 Apr 2020, 09:42
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
Source: Jeff Sackman questions - https://www.gmathacks.com
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - https://www.gmathacks.com
Probability = 7C2 * 3! / 8C3 * 3! = 7*6*3/8*7*6 = 3/8
IMO B
Re: A certain class consists of 8 students, including Kim. Each day, three
[#permalink]
12 May 2021, 07:12
12 May 2021, 07:12
Bowtie wrote:
I agree with B 3/8
first task = 1/8 chance
second task= 7/8*1/7 = 1/8
third task = 7/8*6/7*1/6 = 1/8
chance that she gets picked for one of these is 1/8+1/8+1/8 or 3/8
first task = 1/8 chance
second task= 7/8*1/7 = 1/8
third task = 7/8*6/7*1/6 = 1/8
chance that she gets picked for one of these is 1/8+1/8+1/8 or 3/8
Hi, can you explain the reason behind 7/8 and 7/8*6/7?
Re: A certain class consists of 8 students, including Kim. Each day, three
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05 Jan 2024, 23:49
05 Jan 2024, 23:49
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