The correct answer is:
A. P(both even) = [tex]\frac{(_{4}P_{1})(_{3}P_{1})}{_{9}P_{2}}[/tex]
The expression would determine the probability that both digits are even.
[tex]|Huntrw6|[/tex]
The expression which would determine the probability that both digits are even which is required for bicycle lock is (4P1)(3P1)/(9P2).
The permutation is the arrangement of the things or object in a systematic order, in all the possible ways. The order of arrangement in permutation is in linear.
A bicycle lock requires a two-digit code of numbers 1 through 9, and any digit may be used only once. The probability of choosing 2 digits from 9 is,
[tex]^9P_2[/tex]
There are total 4 even numbers {2,4,6,8}. The probability of choosing first digit's even from 4 even numbers is,
[tex]^4P_1[/tex]
For the second digit to be even is,
[tex]^3P_1[/tex]
Thus, the favorable outcome is, [tex](^4P_1)(^3P_1)[/tex] and total outcome is [tex]^9P_2[/tex]. Thus, the expression which would determine the probability that both digits are even is,
[tex]P=\dfrac{(^4P_1)(^3P_1)}{^9P_2}[/tex]
Thus, the expression which would determine the probability that both digits are even which is required for bicycle lock is (4P1)(3P1)/(9P2).
Learn more about the permutations here;
https://brainly.com/question/12468032
#SPJ3
