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: Calculate the total number of two - digit codes

The formula for permutations \(_{n}P_{r}=\frac{n!}{(n - r)!}\). Here, \(n = 9\) (digits from \(1\) to \(9\)) and \(r=2\). The total number of two - digit codes (where order matters) is \(_{9}P_{2}=\frac{9!}{(9 - 2)!}=\frac{9!}{7!}=9\times8 = 72\).

Step2: Calculate the number of two - digit even codes

The even digits from \(1\) to \(9\) are \(2\), \(4\), \(6\), \(8\) (so \(n = 4\)). For the first digit (even), we have \(_{4}P_{1}=\frac{4!}{(4 - 1)!}=4\) choices. After choosing the first even digit, for the second digit (even, since no repetition), we have \(_{3}P_{1}=\frac{3!}{(3 - 1)!}=3\) choices. The number of two - digit even codes (where order matters) is \((_{4}P_{1})(_{3}P_{1})=4\times3 = 12\)

Step3: Use the probability formula

The probability formula for permutations (since the code is an ordered arrangement) is \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P(\text{both even})=\frac{(_{4}P_{1})(_{3}P_{1})}{_{9}P_{2}}\)

Answer:

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