A class contains five juniors and five seniors. If one member of the

Question

A class contains five juniors and five seniors. If one member of the class is assigned at random to present a paper on a certain subject, and another member of the class is randomly assigned to assist him, what is the probability that both will be juniors?

A. 1/10
B. 1/5
C. 2/9
D. 2/5
E. 1/2

Kudos for a correct solution.

A. 1/10
B. 1/5
C. 2/9
D. 2/5
E. 1/2

Kudos for a correct  solution .

noTh1ng · Answer

I might be wrong:
but what is the matter with

P(junior) * P(Junior) = 5/10 * 4/9 = 2/9 ?

AbdurRakib · Answer

This is an interesting probability problem. Please   correct me if I am wrong

The probability of first assigned student as Junior =Number of Juniors /Total Students =5/10= 1/2

The probability of Second assigned student as Junior = Number of remaining junior students/Total remaining Students= 4/9

So the probablity of both students assigned as Junior=(1/2) * (4/9)= 2/9

So the  Correct Answer C

Bowtye · Answer

I think you're right. The other way to do it (but the long way) is to figure out the probability that it is not two Juniors.

2 seniors = P(Senior) * P(Senior) = 2/9

1 Senior and 1 Junior = (1/2) *(5/9)*2 = 5/9

Probability that it is not two Juniors is 5/9+2/9 = 7/9 so the probability that it is two juniors is 1- (7/9) = 2/9.

Again unnecessarily long but it does check out

minwoswoh · Answer

Hi Wofford09,

I think your solution is correct. However, and as you stated, I don´t think this is the fastest approach.
This is my point: what you are doing follows the reasoning of "Desired outcome = 1 - Undesired outcome". This approach is a shortcut when the undesired outcome is easier to calculate than the desired. In this case, it is easier to get to the desired outcome with the simple calculation (1/2)(4/9) = 2/9.

Nonetheless, it´s great to know both methods because they will come in handy at 700+ problems. So thanks for pointing it out!

Cheers,

=)

noTh1ng · Answer

Apparently the solution is 2/5?

Bunuel  can you help?

Bowtye · Answer

I would like to see an explanation as well. I just googled to see if anyone else had an issue with this question and Kaplan shows the solution as 2/9 in their ACT guide book.

Thanks

Bunuel · Answer

The OA is 2/9. Edited.

kunal555 · Answer

number of ways of selecting 2 members our of 10 members - 10C2
number of ways of selecting 2 juniors out of 5 - 5C2
probability that both members are juniors is

\frac{5C2}{10C2}

or  \frac{5*4}{10*9}

or  \frac{2}{9}

Answer:- C

ravigupta2912 · Answer

I solved it as under:-

Total ways of picking 2 students out of 10 = 10*9 = 90
Total ways of picking 2 juniors only = 1C5 * 1C4 = 20

P (2 juniors only) = 20/90 = 2/9

I'm very weak in P&C, so took the long route. But yeah, one can also just do 2C5/2C10 =10/45 = 2/9

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A class contains five juniors and five seniors. If one member of the

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