I have been left stumped by the following question:

  1. From a group of $6$ girls and $5$ boys, a committee of four is selected. If a committee is selected at random find the probabilty that there is/are:

e. a majority of girls

I attempted to solve this question by

$$\frac{{6\choose 3}\times{8\choose 1}}{ {11\choose 4}}$$

taking how many ways you could choose $3$ girls from $6$ $6\choose 3$ and took that number and multiplied it by how many ways you could choose the final committee member $8\choose 1$ which gives you $20 \times 8 = 160$. I then took this number and and divided it by the total number of possible different committees ${11\choose 4} = 330$. This netted me the incorrect answer, the correct solution was as follows:

$$\frac{{6\choose 3} \times {5\choose 1} + {6 \choose 4}}{330}$$

Looking at this solution it is similar to mine but it splits up the problem into two "cases", the first is where you choose $3$ girls then $1$ boy, the other when it chooses all girls. I can see the logic in splitting up the cases but can anyone point me to a resource that gets more in depth in when you need to use cases? Or more likely am I missing a basic rule? The teacher was unable to explain when and why to use cases so any guidance would be greatly appreciated :)

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    $\begingroup$ It's not a matter of "using" cases as much as it is a matter of there are cases. Committees with a majority of girls naturally fall into one of two cases - that's a fact and an observation, not a choice we make. How do you make this observation? Well, just by paying attention and thinking about it, there's no rule for that. By the way, your method overcounts, because it e.g. considers choosing {Alice, Bryanna, Coral} and then {Deni} as distinct from choosing {Alice, Bryana, Dani} and then {Coral}, even though those are both the same committee. $\endgroup$ – arctic tern May 10 '17 at 8:19
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    $\begingroup$ Just to reiterate the above in slightly different words; how can you rephrase 'a majority of girls' into something we can work with? It means the committee must have 3 girls and one boy or all girls. The word 'or' in probability is most often an addition. If you roll a die, what's the probability you get a two or a three? $\frac 1 6+\frac 1 6$. $\endgroup$ – Arby May 10 '17 at 8:26

The problem with your strategy is that when you pick 3 girls from 6 and then the last member at random among 8, you can reach any committee of four girls by four different ways: pick A,B,C and then D, or A,B,D and then C, etc.

Hence these all-girls committees are counted four times in your combinations...

  • $\begingroup$ Using 6C3 counts all the different combinations where order doesn't matter, but because instead of using the two cases and adding them, I multiply them by the remaining 8C1 and thus make order matter. With 6C3 * 5C1 it is okay to multiply because I am not choosing from my original pool of 6 and am thus not "double dipping" would this be a correct way of attempting to understand this? $\endgroup$ – Oliver Chalk May 10 '17 at 8:36
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    $\begingroup$ I wrote down some combinations on paper and see what you mean about doubling up, you can have the following: G1, G2, G3 + G4 G1, G3, G4 + G2 Where the girl chosen after the plus is the one you chose from the 8 at random. $\endgroup$ – Oliver Chalk May 10 '17 at 8:53

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