Hard ACT Math Probability Problem with Solution

Hard ACT Math Probability Problem with Solution

Today I will give a solution to yesterday’s ACT math probability problem. Here is the question again followed by a ful solution.

Level 5 Probability

Suppose that x will be randomly selected from the set

{–3/2, –1, –1/2, 0, 5/2}

and that y will be randomly selected from the set

{–3/4, –1/4, 2, 11/4}.

What is the probability that x/y < 0 ?

A. 1/100
B. 1/20
C. 3/20
D. 1/3
E. 2/5

Solution: There are 5 possible values for x and 4 possible values for y. By the counting principle, there are 5 ⋅ 4 = 20 possibilities for xy (possibly with some repetition).

In order for x/y < 0 to be true, x and y need to be opposite in sign. The number of ways for this to happen (possibly with repetition) is

3 ⋅ 2 + 1 ⋅ 2 = 6 + 2 = 8

So the desired probability is 8/20 = 2/5, choice E.

Notes: (1) In this question when we mention the possibilities for xy, we are ignoring the fact that there could be repeated values. For example, we have (0)(–3/4) = 0 and (0)(–1/4) = 0. Technically these two computations give a single value for xy. Nonetheless, to compute the desired probability we need to think of these as distinct possibilities.

(2) There are 3 negative values for x and 2 positive values for y. This gives us 3 ⋅ 2 = 6 negative values for xy.

There is 1 positive value for x and 2 negative values for y. This gives us 1 ⋅ 2 = 2 more negative values for xy.

So altogether there are 6 + 2 = 8 possibilities that lead to a negative value for xy.