Probability and Expected Value on the ACT® Math Test: Full Guide

Consider this Example

You have the following marbles in a bag:

• 5 black
• 3 blue
• 6 red
• 10 yellow

You reach your hand in. What is the probability that you draw NOT yellow? In this question, we need to know the desired outcome and the total outcomes. "Desired outcome" is what it sounds like: what do we want to do? We want to draw NOT yellow. The sum of NOT yellow is 5 black + 3 blue + 6 red. Drawing any of those 14 marbles would qualify as "NOT yellow."

Then we need to know the total outcomes. Total outcome is the total possibility of what could occur. When we reach our hand in the bag, we could draw any marble. There are 5 black + 3 blue + 6 red + 10 yellow marbles. We could draw any of those 24 marbles.

Then the probability of drawing NOT yellow, our desired over our total outcomes is 14/24.

Independent Probability

Probability gets a little, but not much, trickier when you have to deal with dependent probability. You're probably reciting the phrase you learned in school, "independent means it depends on what happened before." But what does that mean?

If I want to know the probability not of drawing one yellow marble, but of drawing a yellow marble twice in a row, then I have to use dependent probability. That means I need to know the probability of the first draw, then the probability of the second draw, then the probability of both of those draws happening.

Problems with dependent probability usually involve basic arithmetic. For example, if you have the probability of drawing 1 yellow marble, then your desired outcome is 10 yellow marbles over your "total outcomes" of 24 total marbles. 10/24 – which simplifies, but we can worry about that after we do the math.

Then, we need to find the probability of drawing a yellow marble. This might sound familiar because it's what we just solved above. The ACT® will make it painfully obvious whether or not there is replacement. They will tell you whether or not you put the first marble you draw back in the bag or not. For this example, we are replacing the marble that is, we draw a marble. Replace it. Draw again. Our probability for the second draw is therefore still 10/24.

We have found the probability for both events is 10/24. But the probability of both happening isn't 10/24. The probability of both happening is 10/24 × 10/24 = 100/576. The probability of drawing a yellow, replacing it, then drawing it again is 100/576.

Dependent Probability

Dependent probability likely makes sense given the above example on independent probability. Instead of replacing the marble, we will not replace the marble. Simple as that. So what's the probability of drawing two yellow marbles, given you do not replace the marble that you draw.

The probability of drawing a yellow is 10/24. Then we need to see what the probability of drawing a second yellow without replacement is. Because we're looking for the probability of drawing a yellow first, then a yellow again, we need to find the probability assuming that our first draw pulled a yellow marble out of the bag. 1 yellow marble was already drawn before. So both the number of yellow marbles and the total number of marbles go down by one 9/23.

Then to find the probability of both happening – draw a yellow marble, then draw a second without replacement – is 10/24 × 9/23 = 90/552.

Practice Section

How will this show up on the ACT® Math Section?

The ACT® Math Test will almost certainly give you probability questions. The questions are seldom more difficult than above. The most common error students make is not reading the question. The text of the question will give you vital information about whether there's replacement or not.

Expected Value

You're likely to get an expected value question on the ACT® Test. The thing about expected value that not all students intuitively understand is that it's not necessarily the value you expect to get. Let me explain.

Consider this example

You roll a six-sided die. Each roll gives you a corresponding number of points, i.e. 1 gives you 1 point. What is the average value of the rolls? To know the average value of any given roll, we need to the probability of each value being rolled.

What's the probability of rolling 1?

• 1 1/6
• 2 1/6
• 3 1/6
• 4 1/6
• 5 1/6
• 6 1/6

So how many points do we get for rolling a 1? 1. The value of the roll times its probability gives me 1/6. 2? 2. The value of 2 times its probability is 2/6.

1 1/6

2 2/6

3 3/6

4 4/6

5 5/6

6 6/6

These numbers are all well and good, but what do they tell me? We found how many points any given result 1-6 could give us. We can now find the average of any given roll.

1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6

If we reduce that fraction, we get 3 1/2. The expected value of any given roll is 3.5. How does that make sense? 3.5 isn't even a value you can roll. That's true. But for all rolls over a period of time, the average value of your rolls is 3.5. That's the value you can "expect." It's your expected value.