Proof verification and explanation in probability

Six regimental ties and nine dot ties are hung on a tie holder. Sergio takes two simultaneously and randomly. What is the probability that both ties are regimental?

I have seen that the probability that, not counting the order, the two extracted are between 6 fixed and none of the other 9; therefore if $$E$$ is the event then: $$\text{Pr}[E] = \frac{C_{6,2}\cdot C_{6,0}}{C_{15,2}} = 1/7$$ where $$C_{n,k} = \frac{D_{n,k}}{P_k} = \frac{n!}{k!(n-k)!} = {n \choose k}$$

For some it is more immediate to calculate the result as: $$6/15 \cdot 5/14 = 30/210 = 1/7.$$

Could I please have a clear step-by-step explanation?

• The two are the same, except the second is an ordered picking: "First, pick a regionalmental tie, then pick a regimental tie again from the remaining ties." Hence there are no $1/2$ factors that cancel from numerator and denominator as in the case of the choosing approach. – Alex R. Jun 1 at 20:40
• @AlexR. Thank you very much. Can you put, please, a more simple answer step by step? I had no intuition about these problems, and I admit it. I ask you kindly to explain it to me very simply step by step with an answer. Thank you very much. – Sebastiano Jun 1 at 20:48
• Thank you to David G. Stork for to have edited my question. – Sebastiano Jun 1 at 21:02

Even though he takes the ties simultaneously, we can still calculate it as if he takes one at a time randomly without replacing the first tie.

At the beginning, there are $$15$$ ties, and $$6$$ of them are regimental. So the probability of taking a regimental tie is $$\frac{6}{15}$$.

For the second tie, there are only $$14$$ left (since he already took $$1$$), and if he took a regimental tie on the first pick, there are $$5$$ regimental ties left. So the probability of taking a regimental tie as the second tie is $$\frac{5}{14}$$.

To find the probability of both these events happening, just multiply the individual probabilities. $$\frac{6}{15}\times\frac{5}{14}=\frac{1}{7}$$ So the final answer is $$\frac{1}{7}$$.

It uses conditional probabilities: if you denote $$A_1$$ the event ‘the first tie is regimental tie and $$A_2$$: the second tie is also a regimental tie, you seek for $$P(A_1\cap A_2)$$. Now, as $$P(A_2 \mid A_1)=\frac{P(A_1\cap A_2)}{P(A_1)},$$ you can rewrite it as $$P(A_1\cap A_2)= P(A_1)P(A_2\mid A_1).$$ Edit:

The first method, as I understand it, counts the number of favourable cases (the regimental ties are $$2$$, chosen among $$6$$ and none among the $$9$$ bow ties, and divides by the total number of possible cases (any two ties among atotal of $$15$$).

• And how do I get the expression of the combinations using binomial coefficients? Bernard I don't remember anything :-(. Can you complete it, please? – Sebastiano Jun 1 at 21:04
• @Sebastiano: I've added a possible explanation, but don't ssee the usefulness of $C_{6,0}$ (which, incidentally, should be $C_{9,0}$, I think). – Bernard Jun 1 at 21:17
• I have undertsood your question well...but excuse very much but...I have given the check mark to the user with a low score. Alwayssssssssssssssssssssssss thank youuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu. – Sebastiano Jun 1 at 21:22
• @Sebastiano; That's quite unimportant for me. You did well! – Bernard Jun 1 at 21:24