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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(both even) = 6P1)GP1)
9P2
P(both even) = 65G(1)
9C2
P(both even) = (SP:/GP:)
P2
P(both even) = (C1)GC)
gC₂
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Answer:

₄P₁ ₃P₁ / ₉P₂

Step-by-step explanation:

The order of the digits is important, so we need to use permutations.  There are 4 even digits between 1 and 9 (2, 4, 6, and 8).

The number of ways to select the first even digit from 4 is ₄P₁.

The number of ways to select the second even digit from 3 is ₃P₁.

The number of ways to select 2 digits from 9 is ₉P₂.

The probability is therefore ₄P₁ ₃P₁ / ₉P₂.

Answer:

A: p (both even) = (4p1)(3p1)/9p2

Step-by-step explanation:

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