From an urn containing $n$ balls numbered from $1$ to $n$, $3$ balls are extracted randomly (without reinsertion). Assuming there is concordance to the $k-th$ extraction if in this extraction the ball with number $k$ comes out, calculate the probability to have $3$ concordances,
Solution. Placing $A_k =$ {there is concordance to the $k-th$ extraction}, from the law of compound probability follows that the required probability in is given by
$P (A_1 ∩ A_2 ∩ A_3) = P (A_1) P (A_2 | A_1) P (A_3 | A_1 ∩ A_2) = 1/
n · 1/(n – 1) · 1/(n – 2)$
I don't understand why $P(A_1)=1/n$, is it because you have one possibility out of $n$ that the first ball is $1$?
I don't understand why $P (A_2 | A_1)=1/(n-1)$ and $P (A_3 | A_1 ∩ A_2)=1/(n-2)$
Best Answer
As suggested in comment I just re-post my comment, more properly, as an answer.
The answer to you question on the reason why $$P(A_1) = \frac{1}{n}$$ is on the affirmative.
Also, $$P(A_2|A_1)=\frac{1}{n−1}$$ because if you had concordance in the first extraction, then ball number $2$ is still inside the urn (among the other $n-1$ left after extraction) and you have a probability of $1/(n−1)$ to get it.
Similar reasoning applies to the other conditional probability.