If a number is to be randomly selected from set A above and a number

Official Solution:


\(A = \{-10, \ -9, \ -8, \ -7, \ -6, \ -5, \ -4\}\)

\(B = \{-\frac{1}{2}, \ -\frac{1}{3}, \ -\frac{1}{5}, \ 0, \ \frac{1}{5}, \ \frac{1}{3}, \ \frac{1}{2}\}\)

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?


A. \(\frac{8}{49}\)
B. \(\frac{12}{49}\)
C. \(\frac{13}{49}\)
D. \(\frac{18}{49}\)
E. \(\frac{21}{49}\)


For the product to be a negative integer the numbers chosen must have opposite signs and their product must result in an integer.

All number in A are negative, so from B we can only choose positive numbers.

If from B we choose \(\frac{1}{5}\), then from A we can choose -10 or -5. Two cases.

If from B we choose \(\frac{1}{3}\), then from A we can choose -9 or -6. Two cases.

If from B we choose \(\frac{1}{2}\), then from A we can choose -10, -8, -6, or -4. Four cases.

So, total of eight pairs will give a negative integer.

The total number of pairs is \(7*7=49\), thus the probability is \(\frac{8}{49}\).


Answer: A