Question

A woman has eight keys on a key ring, one of which fits the door she wants to unlock. Sherandomly selects a key and tries it. If it does not unlock the door, she randomly selects anotherkey from those remaining and tries to unlock the door with it. She continues in this manner untilthe door is unlocked. LetXbe the number of keys she tries before unlocking the door, countingthe key that actually worked. Find the expected number of keys she will try before she opensthe door.

Answer 1

The expected number of keys she will try before she opens the door is 4.5.

Explanation:

Probability of opening with the first key:

One key out of 8 working, so: 1/8 probability of opening with the first key, that is:

`P(X = 1) = (1)/(8)`

Probability of opening with the second key:

Doesn't work on the first(7/8 probability), works on the second(1/7 probability). So

`P(X = 2) = (7)/(8)*(1)/(7) = (1)/(8)`

Following the logic:

For each value of x from 1 to 8, we have that
`P(X = x) = (1)/(8)`

Find the expected number of keys she will try before she opens the door.

Each outcome multiplied by its probability. So

`E(X) = 1(1)/(8) + 2(1)/(8) + 3(1)/(8) + 4(1)/(8) + 5(1)/(8) + 6(1)/(8) + 7(1)/(8) + 8(1)/(8) = (1+2+3+4+5+6+7+8)/(8) = 4.5`

The expected number of keys she will try before she opens the door is 4.5.