The membership of a committee consists of 3 English teachers

Hello Bunuel,

Can you please explain why you have used combinations instead of permutations to calculate the probability?

Thanks in advance,

Hi sarafbhushan,

The combinatioan has to be used due to the below reason.

Let us say if we have 3 people A, B and C. If we are asked to select a couple. we can select AB/BC/CA: Please mnote that selecting AB is same as selecting BA as my couple is same in both the cases. Same is the case with selection teachers for a committee, as is the case in the question.

3C2 = 3C1 (not selecting 1) = 3 (AB, BC, CA)
3P2 = 6 (If Ab is not = BA)

Hope it is clear

As explained above, order of the members in the committee is not important. For example, \(C^2_9\) gives the # of all committees of 2 we can choose out of 9 teachers (a, b, c, d, e, f, g, h, i): (a, b), (a,c), (a,d), ... Now, committee (a,b) is the same committee as (b,a), so that's why we should use C (which counts (a,b) only once) instead of P (which counts (a,b) as well (b,a)).

Check Probability and Combinations chapters of Math Book for more on this issues (link in my signature).

Hope it's clear.

Hi all,

I was trying to solve this problem in reverse way and getting wrong answer. Im solving as “1 - probability of selecting both from maths and social” that is 1-6c2/9c2 = 7/12. Anyone explain where I went wrong .

The opposite event of both being English teachers is {both are Mathematics and Social teachers} + {one is English teacher and one is Mathematics or Social teacher}.

\(P=1 - (\frac{C^2_6}{C^2_9}+\frac{C^1_3*C^1_6}{C^2_9})=1-(\frac{15}{36}+\frac{18}{36})=\frac{1}{12}\)

Or: \(P=1-(\frac{6}{9}*\frac{5}{8}+2*\frac{3}{9}*\frac{6}{8})=\frac{1}{12}\) (we multiply 3/9*6/8 by 2 because English/Not English can occur in two ways: {English, Not English} and {Not English, English}).

Answer: D.

Hope it's clear.

What would be the probability that we don't have any English teacher?
6/9*5/8=30/72 is clearly wrong, bc 1-30/72 is not 1/12.

Also, what does C^2_3 mean? we didn't have it in high school I think

The probability that we don't have any English teacher is 6/9*5/8 = 30/72 = 5/12.

If you want to get this by 1 - (the probability of the opposite event), then you should do:

P(no English teachers) = 1 - (P(both are English teachers) + P(one of the two is an English teachers)) =

= 1 - (1/12 + 3/9*6/8*2) =

= 5/12.

P(one of the two is an English teachers) = P(the first one is an English teacher and the second one is not) + P(the first one is not an English teacher and the second one is an English teacher) = 3/9*6/8 + 6/9*3/8 = 3/9*6/8*2.

Now, about C. C stands for combinations: \(C^k_n=\frac{n!}{k!(n-k)!}\) and this gives the number of ways we can choose k objects out of n distinct objects. For example, the number of ways to choose 2 teachers out of 9 is \(C^2_9=\frac{9!}{2!(9-2)!}=\frac{9!}{2!7!}=36\) (note that sometimes it's written as \(C^2_9\) or as \(C^9_2\) or as \(9C2\) but it still means choosing 2 out of 9 and equals to \(\frac{9!}{2!(9-2)!}=\frac{9!}{2!7!}=36\) ). So, we can make 36 different committees of 2 out of 9 teachers.

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Hope it helps.

­If you were to use permuations in calculating both the favorable and the total, you'd get the same answer, but using combinations makes more sense since the order of elements doesn't matter when you form committees in which there's no difference between the positions in the committees.­