Direct Method

2 Cases:

1. When x is multiple of 4 (y can be anything) and when y is multiple of 4 (x can be anything) (Note : Exclude cases where both are div by 4)
2. When x and y are multiples of 2 (and not 4 which is covered in prev cases)

Case 1: (5/20)(1) * (1)(5/20) - (1/4)(1/4) = 1/2 - 1/16
Case 2: 5/20*5/20 = 1/16

Adding: [spoiler]1/2[/spoiler]

knight247 wrote:If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?
A. 1/4
B. 1/3
C. 3/8
D. 7/16
E. ½

OA is E. Fairly simple to solve using the complementary method. Would like some help with the direct method. No Complementary Please.

Case 1: even*even = multiple of 4.
1/2 the numbers in each range are even.
P(x is even AND y is even) = 1/2 * 1/2 = 1/4.

Case 2: multiple of 4 * odd = multiple of 4.
Since every other even number is a multiple of 4, and 1/2 the numbers in each range are even, 1/4 of the numbers in each range are multiples of 4.
1/2 the numbers in each range are odd.
Thus:
P(x is a multiple of 4 AND y is odd) = 1/4 * 1/2 = 1/8.
P(x is odd AND y is a multiple of 4) = 1/2 * 1/4 = 1/8.

Since getting the first result (1/4) OR the second result (1/8) OR the third result (1/8) is a good outcome, we add the fractions.
1/4 + 1/8 + 1/8 = 1/2.

The correct answer is E.

GMATGuruNY wrote:

knight247 wrote:If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?
A. 1/4
B. 1/3
C. 3/8
D. 7/16
E. ½

OA is E. Fairly simple to solve using the complementary method. Would like some help with the direct method. No Complementary Please.

Case 1: even*even = multiple of 4.
1/2 the numbers in each range are even.
P(x is even AND y is even) = 1/2 * 1/2 = 1/4.

Case 2: multiple of 4 * odd = multiple of 4.
Since every other even number is a multiple of 4, and 1/2 the numbers in each range are even, 1/4 of the numbers in each range are multiples of 4.
1/2 the numbers in each range are odd.
Thus:
P(x is a multiple of 4 AND y is odd) = 1/4 * 1/2 = 1/8.
P(x is odd AND y is a multiple of 4) = 1/2 * 1/4 = 1/8.

Since getting the first result (1/4) OR the second result (1/8) OR the third result (1/8) is a good outcome, we add the fractions.
1/4 + 1/8 + 1/8 = 1/2.

The correct answer is E.

I understood Case I:
even x even = multiple of 4
Prob. of choosing an even number from 1-20 x Prob. of choosing an even number from 21-40 = 1/2 * 1/2 = 1/4

I have a doubt in Case II:
For xy to be multiple of 4 -
a) we need to get multiple of 4 from 1-20 and multiply with any number between 21-40
= 1/4
b) we just need to get multiple of 4 from 21-40 and multiply with any number between 1-20
= 1/4

Total Prob = 1/4 + 1/4 + 1/4 = 3/4

Why do restrict over selves to "P(x is a multiple of 4 AND y is odd)" and "P(x is odd AND y is a multiple of 4)"
I feel once x is multiple of 4, y can be any number between 21-40 and once y is multiple of 4, x can be any number between 1-20

@gautam
In case two we have to consider Multiple of 4*ODD and Vice-Versa because
Multiple of 4*Any Number also includes Multiple of 4*Even i.e. Even*Even which is already considered in case 1.

yes perfect the answer comes out to be 1/2.