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A woman has eight keys on a key ring, one of which fits the door she wants to unlock. She randomly selects a key and tries it. If it does not unlock the door, she randomly selects another key from those remaining and tries to unlock the door with it. She continues in this manner until the door is unlocked. Let \( X \) be the number of keys she tries before unlocking the door, including the key that actually works. Find the expected number of keys she will try before she opens the door.

Answer :

Answer:

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

Step-by-step 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:

[tex]P(X = 1) = \frac{1}{8}[/tex]

Probability of opening with the second key:

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

[tex]P(X = 2) = \frac{7}{8}*\frac{1}{7} = \frac{1}{8}[/tex]

Following the logic:

For each value of x from 1 to 8, we have that [tex]P(X = x) = \frac{1}{8}[/tex]

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

Each outcome multiplied by its probability. So

[tex]E(X) = 1\frac{1}{8} + 2\frac{1}{8} + 3\frac{1}{8} + 4\frac{1}{8} + 5\frac{1}{8} + 6\frac{1}{8} + 7\frac{1}{8} + 8\frac{1}{8} = \frac{1+2+3+4+5+6+7+8}{8} = 4.5[/tex]

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

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