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I was running the morning standup this morning and in contrary to the usual alphabetical order I announced I was going to go in "random" order today, however I then ended up giving all the ladies of the team the word first before 'randomly' choosing the gents.

If my selection was really random what is the probability of having all the ladies come first?

4 ladies 7 gents

My back of the envelope calculation was:

4/11 * 3/10 * 2/9 * 1/8 = 0.003030303
  • Is this anywhere correct?
  • How to write this problem in a formal way?

P.S. This is not a HW question but can be considered as a kind of self study.

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  • $\begingroup$ if this is HW please add the "self-study" tag $\endgroup$ – Antoine Feb 8 '16 at 11:56
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First, note that every ordering of the ladies in the first four positions will give the same event: namely ladies first! Then, note that any ordering of the gentlemen in the last seven positions will also give the event ladies first! How do we account for all these outcomes?

Here are two formal ways to accomplish this. We begin by considering the sample space in the ordered case. By ordered here we mean that we can tell the ladies apart and the same for the gentlemen. You can name them if it makes things easier.

Assuming randomness and equilikely events, we have to count the ways this can occur and then divide by the total number of orderings to get our probability. Let's start from the total number of orderings of these eleven people. This is given by the number of permutations of course, $11!$. Then, how many ways are there to order the four ladies in the four positions? Well, the same reasoning will give $4!$ ways. But now for each of these orderings, we can order the gentlemen in $7!$ ways and still have an ordering in which the ladies come first. Thus we may write

$$\Pr \left( \text{Ladies first} \right) = \frac{4!7!}{11!} = 0.0030 $$

and it turns out that your calculation was correct.

Now, an alternative way to view this. Assume now that we are in the unordered case and we cannot tell the ladies apart and the gentelemen either. All we know is that a person is a lady or a gentleman. Is there a way to compute the probability in this case?

Well yes, if we consider the proper sample space. First let's count the total number of unordered arrangements of the ladies and the gentlemen. This is given by the binomial coefficient $\binom{11}{4, 7}$ and the desired outcome now is just one of these arrangements. Hence, we write

$$\Pr \left( \text{Ladies first} \right) = \frac{1}{\binom{11}{4, 7}} = \frac{4! 7!}{11!}$$

and we get of course the same result but from different principles!

Hope this helps.

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