A certain class consists of 8 students, including Kim. Each day, three

Question

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

Bunuel · Accepted Answer

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?

Bunuel · Answer

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.

metallicafan · Answer

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!

metallicafan · Answer

Thank you Bunuel!, I have updated the post with the choices. 8-)

pratikbais · Answer

Thankx Bunuel, the explanation covers all possible ways of solving the problem

Gwydion · Answer

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

Bowtie · Answer

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

msharmita · Answer

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

Bunuel · Answer

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.

Kinshook · Answer

Probability = 7C2 * 3! / 8C3 * 3! = 7*6*3/8*7*6 = 3/8

IMO B

hemanthPA · Answer

P(A) = 1/8
P(B) = 7/8*1/7 = 1/8
P(C) = 7/8*6/7*1/6 = 1/8

P(A)+P(B)+P(C) = 3/8

luisdicampo · Answer

Deconstructing the Question  
3 different students are chosen (A, then B, then C) from 8 total including  Kim . Find  P(	ext{Kim gets at least one task}) .

Key Idea  
Compute the complement:  1 - P(	ext{Kim is never chosen}) .

Step-by-step  
 Not chosen for Task A:   7/8  (7 of the 8 students are not Kim)

Not chosen for Task B:  After A is assigned to someone else,  7  students remain and Kim is still among them.
So, Kim not chosen for B:  6/7

Not chosen for Task C:  After A and B go to others,  6  students remain and Kim is still among them.
So, Kim not chosen for C:  5/6

P(	ext{Kim never chosen})=(7/8)(6/7)(5/6)=5/8

So,
 P(	ext{Kim chosen at least once})=1-5/8=3/8

Answer:     (B) 3/8

A certain class consists of 8 students, including Kim. Each day, three

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A certain class consists of 8 students, including Kim. Each day, three

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