Proof verification and explanation in probability

Question

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?

Answer 1

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 $\frac{1}{7}$.

Answer 2

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 $15$).