Question

Out of 6 keys (A to F), only 1 key is used to open the front door lock of a house and from the set of another 4 keys (1 to 4), 1 key opens the bedroom. What is the probability that the first key used is C and second key used is not 3?

• A. $\frac{1}{24}$
• B. $\frac{1}{8}$
• C. $\frac{2}{24}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{8}$

Number of keys to open front door = 6

Number of keys to open bedroom door = 4

So, total number of outcomes will be:

Number of possible outcomes = 24

We have to choose first key as C and second key as not 3.

We don't have to choose 3 - it means that we have to choose from either 1, or 2, or 4.

So, the possible outcomes are:


Number of favorable outcomes = 3

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$=\frac{3}{24}=\frac{1}{8}$