# Combinations: Understanding the need for “cases”

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 :)

• 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. – arctic tern May 10 '17 at 8:19
• 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? $\frac 1 6+\frac 1 6$. – Arby May 10 '17 at 8:26