Bunuel wrote:
If four integers, a, b, c, and d are chosen at random, with replacement, what is the probability that ab + cd is even?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
Solution:Since odd + odd = even and even + even = even, then either both ab and cd are odd or they are both even.
If both ab and cd are odd, then a, b, c and d have to be all odd. So the probability is ½ x ½ x ½ x ½ = 1/16 (notice that the probability being odd is ½ and the probability being even is also ½).
If both ab and cd are even, then we have:
1) a = even, b = odd, c = even, d = odd
2) a = even, b = odd, c = odd, d = even
3) a = even, b = odd, c = even, d = even
4) a = odd, b = even, c = even, d = odd
5) a = odd, b = even, c = odd, d = even
6) a = odd, b = even, c = even, d = even
7) a = even, b = even, c = even, d = odd
8) a = even, b = even, c = odd, d = even
9) a = even, b = even, c = even, d = even
The probability of any of the cases above is also 1/16. Therefore, the overall probability is 1/16 + 9 x 1/16 = 10/16 = 5/8.
Alternate Solution:ab + cd is even if both ab and cd is even or both ab and cd is odd. Let’s find P(ab is even) (which equals P(cd is even)) and P(ab is odd) (which equals P(cd is odd)):
P(ab is odd) = P(a is odd and b is odd) = 1/2 x 1/2 = 1/4
P(ab is even) = 1 - P(ab is odd) = 1 - 1/4 = 3/4
Now, the probability that both ab and cd are even is found to be P(both ab and cd are even) = P(ab is even) x P(cd is even) = 3/4 x 3/4 = 9/16. Similarly, P(both ab and cd are odd) = P(ab is odd) x P(cd is odd) = 1/4 x 1/4 = 1/16. Thus, P(ab + cd is even) = P(both ab and cd are even) + P(both ab and cd are odd) = 9/16 + 1/16 = 10/16 = 5/8.
Answer: D