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A certain class consists of 8 students, including Kim

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A certain class consists of 8 students, including Kim [#permalink] New post 17 Feb 2012, 06:50
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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

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 - http://www.gmathacks.com
[Reveal] Spoiler: OA

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Re: A certain class consists of 8 students [#permalink] New post 17 Feb 2012, 06:55
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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 [#permalink] New post 17 Feb 2012, 07:12
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 [#permalink] New post 17 Feb 2012, 07:23
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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 [#permalink] New post 17 Feb 2012, 07:49
Thank you Bunuel!, I have updated the post with the choices. 8-)
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Re: A certain class consists of 8 students [#permalink] New post 17 Feb 2012, 09:00
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 [#permalink] New post 17 Feb 2012, 16:43
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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
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Re: A certain class consists of 8 students, including Kim [#permalink] New post 23 Feb 2012, 09:06
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
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Re: A certain class consists of 8 students, including Kim [#permalink] New post 22 Sep 2013, 09:33
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Re: A certain class consists of 8 students, including Kim [#permalink] New post 03 May 2014, 22:35
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 - http://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
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Re: A certain class consists of 8 students, including Kim [#permalink] New post 04 May 2014, 02:01
Expert's post
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 - http://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

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Hope this helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A certain class consists of 8 students, including Kim   [#permalink] 04 May 2014, 02:01
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