Answer to Question #304891 in Statistics and Probability for ced

Question

A key ring contains four office keys that are identical in appearance, but only one will

open your office door. Suppose you randomly select one key and try it. If it does not

fit, you randomly select one of the three remaining keys. If it does not fit, you

randomly select one of the last two.

(a) List the elements of the sample space S using the letters H and M for “hit” and

“miss”, respectively.

(b)To each sample point in S, assign a value x of the random variable X representing

the number of keys that you try until you find the one that opens the door.

(c) Calculate the values of P(X = x) and display them in a table.

Answer 1

(a) $S = \begin{Bmatrix} H, MH, MMH, MMMH \end{Bmatrix}$

(b) With each sample point we'll associate a number of keys that we try until we find the one that opens the door. In the case of H (i.e the first key fits) X = 1; for MH (first key doesn't fit, the second key fits) X = 2; for MMH X = 3 and for MMMH X = 4.

(c) For the first sample point we can choose any of the four keys, the probability of the right choice: $P(H)=P(X=1)=\frac{1}{4}.$

For the second sample point we choose first an incorrect key (M) and then one of the three remaining keys, $P(MH)=P(X=2)=\frac{3}{4}\cdot \frac{1}{3}=\frac{1}{4}.$

In the same way $P(MMH)=P(X=3)=\frac{3}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{1}{4};$

$P(MMMH)=P(X=4)=\frac{3}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}\cdot 1=\frac{1}{4}.$