Answer:
The answer is the first one.
P(both even) = (4P1)(3P1) / 9P2
The probability of selecting two numbers for a bicycle lock such that both numbers are even is P(both even) = StartFraction (4 P 1) (3 P 1) Over 9 P 2 EndFraction, the correct option is A.
Probability is the measure of the likeliness of an event to happen.
The probability has a range from 0 to 1, 0 denotes uncertainty, and 1 indicates certainty.
A bicycle lock requires a two-digit code of numbers 1 through 9.
Any digit can be used only once.
There are total 9 numbers.
The no. of number that has to be chosen is 2.
The probability that both digits even have to be determined.
The no. of even numbers is 4
The no. of ways for selecting 1 number from the 4 even numbers is ⁴C₁
The no. of ways for selecting the second number when the digits cannot be repeated is given by ³C₁
The total no. of ways by which 2 numbers are selected from 9 numbers is ⁹C₂
The probability is equal to the ratio of the favorable outcomes to the total outcomes.
The probability of selecting two numbers for a bicycle lock such that both numbers are even is
= ⁴C₁ * ³C₁ / ⁹C₂
= 4 * 3 * 2/ (9 * 8)
= 1/3
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