Question

a bicycle lock requires a two - digit code of numbers 1 through 9, and any digit may be used only once. which expression would determine the probability that both digits are even?

• $p(\text{both even}) = \frac{(_{4}p_{1})(_{3}p_{1})}{_{9}p_{2}}$
• $p(\text{both even}) = \frac{(_{4}c_{1})(_{3}c_{1})}{_{9}c_{2}}$
• $p(\text{both even}) = \frac{(_{5}p_{1})(_{4}p_{1})}{_{9}p_{2}}$
• $p(\text{both even}) = \frac{(_{5}c_{1})(_{4}c_{1})}{_{9}c_{2}}$

Explanation:

Step1: Identify even numbers from 1 - 9

Even numbers from 1 to 9 are 2, 4, 6, 8. So there are 4 even numbers.

Step2: Determine permutations for favorable and total cases

• For the favorable cases (both digits even): The first digit can be chosen in \(_{4}P_{1}\) ways (permutation of 4 even numbers taken 1 at a time), and after choosing the first even digit, the second digit can be chosen in \(_{3}P_{1}\) ways (permutation of remaining 3 even numbers taken 1 at a time).
• For the total cases: We are choosing 2 - digit code from 9 numbers (1 - 9) with no repetition, so the total number of permutations is \(_{9}P_{2}\) (permutation of 9 numbers taken 2 at a time).
• Probability is the ratio of favorable permutations to total permutations. So \(P(\text{both even})=\frac{(_{4}P_{1})(_{3}P_{1})}{_{9}P_{2}}\)

Answer:

\(P(\text{both even})=\frac{(_{4}P_{1})(_{3}P_{1})}{_{9}P_{2}}\) (the first option)