Answer to Question #118762 in Statistics and Probability for Iqra

Question

Can you please answer my question?
Paula has three keys on a key ring, just one of which opens the door to her
house. As she approaches the door, she selects one key after another, at
random without replacement, trying each key in the door until she finds the
correct key.
(i) Draw up a probability distribution table for X, the number of keys Paula
tries before she opens the door.
(ii) Calculate the expected number of keys that she will try before opening
the door.

Answer 1

Paula has 3 keys on a key ring, 1 of which opens the door to her house. She selects one key after another, at random without replacement.

X = the number of keys Paula tries before she opens the door

Since, she tries the keys without replacement, the value of X = 0, 1 and 2.

Now, P(X = 0) = she opens the door with the first key she tries = 1/3

P(X = 1) = she fails to open with the first key and opens with the second key = 2/3 X 1/2 = 1/3 [since, after failing with the first key there remains 2 keys, so the probability of opening with the second key becomes 1/2]

P(X = 2) = she fails to open with the first key and fails to open with the second key and opens with the third key = 2/3 X 1/2 X 1 = 1/3 [since, after she fails with the first 2 keys, opening the door with the third key becomes sure event]

(i)

The probability distribution of X is:

X : 0 1 2

P(X = x): 1/3 1/3 1/3

(ii)

E(X) = $\sum$x.P(X = x) = 0 X 1/3 + 1 X 1/3 + 2 X 1/3 = 1/3 + 2/3 = 1

Answer: The expected number of keys that Paula will try before opening the door is 1.