Question

A woman has n keys, of which one will open her door. If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her kth try? What if she does not discard previously tried keys?

Answer 1

Answer:

If she discard previously tried keys the probability is 1/n

If she doesn't discard previously tried keys the probability is:

$\frac{(n-1)^{k} }{n^{k} }$

Step-by-step explanation:

If she discard previously tried keys, the probability of open the door in the 1st, 2nd and 3rd try are:

1st try:   $\frac{1}{n}$

2nd try:  $(\frac{n-1}{n})*(\frac{1}{n-1})=\frac{1}{n}$

3rd try: ($\frac{n-1}{n})*(\frac{n-2}{n-1})*(\frac{1}{n}) =\frac{1}{n}$

Where, for example, the probability of open the door in the 2nd try is the multiplication of the probability that the woman didn't open the door in the 1st try with the probability that she open the door in the 2nd try.

So, the probability that she open the door in the kth try is: 1/n

On the other hand, if she doesn't discard previously tried keys, the probability to open the door in the 1st, 2nd and 3rd try are:

1st try:   $\frac{1}{n}$

2nd try:  $(\frac{n-1}{n})*(\frac{1}{n})=\frac{n-1}{n^2}$

3rd try: ($\frac{n-1}{n})*(\frac{n-1}{n})*(\frac{1}{n}) =\frac{(n-1)^2}{n^3}$

So, the probability that she open the door in the kth try is:

$\frac{(n-1)^{k-1}}{n^k}$