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Probability problem

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by AndyMann » Sat Jul 20, 2013 4:54 pm
I have a question regarding the correct method to solve this Kaplan problem:

A 10-member student leadership committee consists of juniors and seniors. There are 4 junior and 6 senior students. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

A 1/10
B 3/10
C 3/5
D 6/7
E 13/14

OA: E

My method ->Since atleast 3 have to be Seniors, (Seniors-S, Junior-J)
Probability of 3S and 3J -> (3/6*3/4)
Probability of 4S and 2J -> (4/6*2/4)
Probability of 5S and 1J -> (5/6*1/4)
Probability of 6S and 0J -> (6/6*0/4)

Net Probability, Summing all the above 9/24+8/24+5/24+0=22/24= 11/12

But the answer is [spoiler]13/14[/spoiler]

I am able to solve the problem using combinations correctly.

Is it possible to solve the above problem using the method mentioned above?


Thanks,
Andy
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by [email protected] » Sat Jul 20, 2013 6:30 pm
Hi Andy,

This question is based on a complex mix of combination and probability. Here's how you solve it:

There are 6 seniors + 4 juniors = 10 total people. We're going to pick groups of 6 under certain conditions.

First, how many total ways are there to pick 6 people from this group of 10:

10!/(6!)(4!) = 210 possible groups of 6

Now, we need to calculate each of the groups that you named:

3 seniors, 3 juniors = (6c3)(4c3) = 80 different groups
4 seniors, 2 juniors = (6c4)(4c2) = 90 different groups
5 seniors, 1 junior = (6c5)(4c1) = 24 different groups
6 seniors, 0 juniors = (6c6)(1) = 1 group

Total groups with at least 3 seniors = 80 + 90 + 24 + 1 = 195

195/210 = 13/14

Final Answer: E

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by vishugogo » Sun Jul 21, 2013 3:50 am
Another method

P(4J and 2S) = 6c2/10c6
= 5/210

1-(5/210) = 195/210
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by GMATGuruNY » Sun Jul 21, 2013 4:07 am
AndyMann wrote: A 10-member student leadership committee consists of juniors and seniors. There are 4 junior and 6 senior students. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

A 1/10
B 3/10
C 3/5
D 6/7
E 13/14

OA: E
TOTAL number of ways to select 6 students:
Number of options for the first student selected = 10. (Any of the 10 students.)
Number of options for the second student selected = 9. (Any of the 9 remaining students.)
Number of options for the third student selected = 8. (Any of the 6 remaining students.)
Number of options for the fourth student selected = 7. (Any of the 9 remaining students.)
Number of options for the fifth student selected = 6. (Any of the 6 remaining students.)
Number of options for the last student selected = 5. (Any of the 5 remaining students.)
To combine these options, we multiply:
10*9*8*7*6*5.

Since the ORDER of the selections doesn't matter -- ABCDEF is the same combination of students as BADCFE -- we divide by the number of ways the 6 selections can be ARRANGED (6!):
(10*9*8*7*6*5)/(6*5*4*3*2*1) = 210.

P(at least 3 seniors) = 1 - P(fewer than 3 seniors).

Total ways to select fewer than 3 seniors:
From the 4 juniors and 6 seniors, there is ONLY ONE WAY to form a combination of 6 with fewer than 3 seniors:
ALL 4 JUNIORS are combined with 2 OF THE 6 SENIORS.
Here, we must select 2 of the 6 seniors to join the 4 juniors.

Number of options for the first senior student selected = 6. (Any of the 6 seniors.)
Number of options for the second senior selected = 5. (Any of the 5 remaining seniors.)
To combine these options, we multiply:
6*5.
Since the ORDER of the selections doesn't matter -- AB is the same combination of seniors as BA -- we divide by the number of ways the 2 selections can be ARRANGED (2!):
(6*5)/(2*1) = 15.

Thus:
P(fewer than 3 seniors) = 15/210 = 1/14.
Thus:
P(at least 3 seniors) = 1 - 1/14 = 13/14.

The correct answer is E.
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