A number x is chosen at random from the set {1, 2, 3, 4} and a number y is chosen at random from the set {5, 6, 7}, independently. What is the probability that the product xy is even?

Introduction / Context:
This question involves basic probability with simple sets of integers. We select one number from each of two sets and evaluate the product. The goal is to find the probability that the product is even, which depends on whether at least one of the factors is even.

Given Data / Assumptions:

• x is chosen uniformly from {1, 2, 3, 4}.
• y is chosen uniformly from {5, 6, 7}.
• The selections of x and y are independent.
• We are interested in P(xy is even).

Concept / Approach:
A product of two integers is even if at least one of the factors is even. Therefore, instead of checking all products directly, we can reason about parities. We can either count favourable outcomes directly or find the complement probability that the product is odd and then subtract from 1.

Step-by-Step Solution:
The set for x is {1, 2, 3, 4}. Here, 1 and 3 are odd, 2 and 4 are even.The set for y is {5, 6, 7}. Here, 5 and 7 are odd, 6 is even.Total number of ordered pairs (x, y) = 4 * 3 = 12.The product xy is odd only when both x and y are odd.Odd x values: 1, 3 (2 choices). Odd y values: 5, 7 (2 choices).Number of odd product pairs = 2 * 2 = 4.Therefore, number of even product pairs = total pairs - odd pairs = 12 - 4 = 8.Probability that xy is even = 8 / 12 = 2 / 3.

Verification / Alternative check:
We can list all 12 ordered pairs and check the product parity. The pairs (1,5), (1,7), (3,5) and (3,7) give odd products; the remaining 8 pairs give even products. This confirms that 8 out of 12 outcomes are favourable, giving the same probability 2/3 when simplified.

Why Other Options Are Wrong:
The value 1/2 would correspond to an equal number of even and odd products, which is not the case here. The fraction 3/4 would require 9 favourable outcomes out of 12, which is too high, while 1/3 would mean only 4 favourable outcomes. The correct count of favourable outcomes is 8, making 2/3 the only consistent choice.

Common Pitfalls:
A common mistake is to focus only on whether x is even and ignore that y can also be even. Another error is to compute the probability of x being even and y being even, rather than at least one being even. Using the complement approach that the product is odd only if both factors are odd is usually the safest and simplest method in such problems.

Final Answer:
The probability that the product xy is even is 2/3.