# What does 'calculate the probability of selecting any given committee' mean?

What does any given committee mean?

How to solve this?

Problem: A committee of size $4$ is to be chosen from a group of $6$ men and $8$ women. The $4$ names are to be chosen randomly “out of a hat”.

What is the probability that at least one woman will be selected?

Thanks

So for the sample space this is my working: $$\binom{6}{4}\binom{8}{4}\binom{14}{4}$$

$6$ women choose $4$ places, $8$ men chooses $4$ places, $14$ people chooses $4$ places.

I guess I don't really understand combinations much.

• Mind showing us what you've tried so far? – platty Aug 3 '17 at 16:38
• Think about how many different sets of 4 different names can be chosen out of a total of 6+8=14 names. Secondly, among 6 male names, how many different sets of 4 different names? Subtract the second quantity from the first. The result is the number of possible committees with at least one woman. Then use that to calculate the probability you are looking for. – Anguepa Aug 3 '17 at 16:42
• Any given committee means a specific group of four people selected from the $14$ available people. – N. F. Taussig Aug 3 '17 at 20:15

## 2 Answers

We shall proceed with complementary counting. The total number of outcomes is $\dbinom{14}{4} = 1001$. The total number of ways where all the chosen people are men is $\dbinom{6}{4} = 15$. Therefore, our answer is $$\frac{1001-15}{1001}=\boxed{\frac{986}{1001}}$$

• Note that if the people were chosen one at a time rather than together, the answer would still be $\dfrac{24024-360}{24024} = \dfrac{1001-15}{1001} = \boxed{\dfrac{986}{1001}}$ – Nairit Aug 3 '17 at 16:47
• Great explanation. But lets say I want 2 men and 2 women in committee, how should I calculate that case? Is it (6 chooses 2)+(8 chooses 2)? – maliiaButterfly Aug 3 '17 at 16:49
• I am a bit confused: Do you mean that there must be 2 men and 2 women in the committee?. If so, the denominator stays the same. For the numerator, we want to choose two men and two women. The number of ways to do this is $\dbinom{6}{2}\dbinom{8}{2}=420$. Therefore, our desired probability is $\dfrac{420}{1001} = \boxed{\dfrac{60}{143}}$ – Nairit Aug 3 '17 at 16:51
• Yes, all the outcomes with 2 men and 2 women. Thank you kindly, Nairit! – maliiaButterfly Aug 3 '17 at 16:53
• @maliiaButterfly It means evaluate "the probability that an unbiased random selection of committee members will fit (the specified pattern)." – Graham Kemp Aug 3 '17 at 22:48

What is the probability of getting a 6 when you roll a dice? It is $\frac 16$ so generally , probability is $\frac{n(E)}{n(S)}$ where set S is the sample space.

• So 1/1001? That's what I thought initially but appeared too simple – maliiaButterfly Aug 3 '17 at 18:08
• No. Maybe my explanation was short . n(E) is not 1 but the number of times a particular event can occur. In the case that committee is to be formed only of men , n(E) =$\dbinom {6}{4}$ = 15 and n(S) = $\dbinom{14}{4}$ = 1001. So , probability that a committee consists of all men is $\frac{n(E)}{n(S)} = \frac {15}{1001}$ Now can you tell what the probability is if the committee consists of all women? – Diaga AoS Aug 3 '17 at 18:23
• So for any given committee means that I gotta list all the possible probabilities? For example, (1 woman, 3 men), (3 women, 1 man) etc.? To answer your question, it is (8 chooses 4)/1001, right? – maliiaButterfly Aug 3 '17 at 18:27
• If there is not a specified case or condition attached then the probability will indeed be 1(It can be calculated by summing the probabilities of all possible committees). But I am guessing the question asks you to calculate the probability of committees with condition attached like committees that consist (1 man , 3 women) etc. – Diaga AoS Aug 3 '17 at 18:40