Probability Questions on ACT Math: Strategies and Practice

Mara is stringing a necklace and she selects each bead at random from a basket of beads. If there are currently 5, yellow beads, 10 red beads, 15 green beads, and 20 blue beads in the basket, what are the chances that she will select a red bead next?

${\desired \outcome}/{\all \possible \outcomes}$

There are 10 red beads, which is our desired outcome. This means 10 is our numerator.

There are also a total of $5 \yellow \beads + 10 \red \beads + 15 \green \beads + 20 \blue \beads = 50 \total \beads$ in the basket. This is our denominator, as it represents all the outcomes possible.

When we put these together, our probability is:

$10/50$

$1/5$

The chances that Mara will select a red bead are 1 in 5 or $1/5$.

What are the odds that Mara will NOT select a green bead?

In order to find a negative probability, we must subtract out the chances that Mara will draw a green bead. (We could also think of this as finding the desired outcome of her selecting a yellow bead, a red bead or a blue bead, which we will cover in more detail in the next section.)

There are only yellow, red, green, and blue beads, so we can add up our odds of yellow, red and blue beads, excluding the green. There are 5 yellow beads, 10 red beads and 20 blue beads, so we can put those together to get our numerator.

$5 + 10 + 20 = 35$

And there are still $5 + 10 + 15 + 20 = 50$ beads total for our denominator.

So what are the odds that Mara will NOT select a green bead?

$35/50$

$7/10$

The odds are 7 in 10 ($7/10$) that Mara will draw any color bead except green.

Instead of asking the odds of Mara selecting only a red bead, what are the odds that Mara will select either a red bead or a green bead if she has 5 yellow beads, 10 red beads, 15 green beads, and 20 blue beads in the basket?

We have increased our odds, since it doesn’t matter whether or not the bead is green or red, so long as the bead we select is NOT blue or yellow (essentially, we are doing another version of our earlier negative problem—”what are the odds that a particular event will NOT happen?”)

This means we can add the probabilities of our individual events together in order to find their combined probability.

So let us find the probability of her drawing a red bead:

$10/(5 + 10 + 15 + 20)$

$10/50$

And let us find the probability of her drawing a green bead:

$15/(5 + 10 + 15 + 25)$

$15/50$

So, if we put the two probabilities together, we’ll have:

$10/50 + 15/50$

$25/50$

$1/2$

Because this problem involves the odds of two events with the same total number of outcomes (there are 50 total possible beads to choose from each time), we could also simply add our two desired outcomes together over the total number of outcomes. So:

$(10 + 15)/(5 + 10 + 15 + 20)$

$25/50$

$1/2$

Either way, the odds of Mara drawing either a red bead or a green bead are 1 in 2, or $1/2$ (50%).

What are the odds that Jenny will roll a pair of dice and get six on both?

A die has six faces, so the odds of rolling any particular number is $1/6$. Because the question is asking us to find the odds of rolling two sixes (and nothing else), we must use our combined probability. So:

$1/6 * 1/6 = 1/36$

There is a 1 in 36 chance that Jenny will roll a pair of dice and get two sixes.

We know that probability is ${\desired \outcome}/{\all \possible \outcomes}$.

Our desired outcome is to get one of the five extra pieces, so our numerator will be 5.

There are 750 puzzle pieces PLUS the extra five pieces in the box total, so our denominator will be: $750 + 5 = 755$

When we put them together, our final probability will be:

$5/755$

Our final answer is D.

We are told that we must find the odds of an event as a ratio of $\in \the 25 - 35 \age \range: \not \in \the 25-35 \age \range$ (in other words, $\desired \outcome: \remaining \outcomes$).

We are given the number of voters in terms of percentages, so we can translate the 42% of voters in the 25-35 age range as $42/100$.

And if the 25-35 age category has a probability of $42/100$, then the remaining voters will have a probability of:

${100 - 42}/100$

$58/100$

Now, we can represent our ratio of $25-35 \voters: \all \other \voters$ as:

$42:58$

Both numbers are divisible by 2, so we can reduce the ratio to:

$21:29$

Our final answer is D.

Now, there are two ways to solve this kind of problem—using proportions or using the strategy of plugging in answers. Let’s look at both methods.

Method 1—Proportions

We are asked to find an additional number of red marbles that we must add to the total number of marbles in order to find a new probability. The current probability of selecting a red marble is:

$12/32$

Now, we are adding a certain number of red marbles and only red marbles. This means that the number of red marbles increases by exactly the same amount that the total increases. We can therefore represent the new probability as:

${12 + x}/{32 + x}$

Now, we want this new probability to be equal to $3/5$, so let us set them up as a proportion.

${12 + x}/{32 + x} = 3/5$

And because this is a proportion, we can cross multiply.

$(32 + x)(3) = (12 + x)(5)$

$96 + 3x = 60 + 5x$

Now solve for $x$.

$36 = 2x$

$18 = x$

So we must add 18 red marbles in order to get a new probability of:

${12 + 18}/{32 + 18$

$30/50$

$3/5$

Our final answer is G, 18.

Method 2—Plugging in answers

The alternative to using proportions is to use PIA. We can simply add the answer options to the 12 red marbles in our numerator and the 32 marbles in our denominator and see which answer choice gives us a final ratio of $3/5$.

Let us begin, as always, with the answer choice in the middle.

Answer option H gives us 28, so let us try adding 28 to both the red marbles and the total number of marbles.

${12 + 28}/{32 + 28}$

$40/60$

$2/3$

This answer is a little bit too large. We can also see that the larger the number we add to both the numerator and the denominator, the larger our probability will be (you can test this by plugging in answer choice J or K—for K, if you add 40 to both 12 and 32, your final probability fraction will be $52/72$ => $13/18$, which is even larger than $2/3$.)

This means that we can eliminate answer choices H, J, and K.

Now let us try answer choice G.

${12 + 18}/{32 + 18}$

$30/50$

$3/5$

We have found our desired ratio.

Our final answer is G, 18.

As you can see, no matter which method you use, you can find the right solution.

Kyle has been tossing a coin and recording the number of heads and tails results. So far, he has tossed the coin 5 times and gotten heads each time. What are the odds that he will get tails on his next coin toss?

You may be tempted to think that our desired outcome (our numerator) is influenced by the number of times Kyle has already tossed the coin and the outcomes so far, but in all actuality, the probability that Kyle will get tails on his next toss is $1/2$.

Why? Because each coin toss is independent of another coin toss. This means that this is a simple matter of determining our desired outcome over the number of total outcomes. There is one possibility of getting tails—numerator 1—and two possible options—heads or tails, denominator 2.

So Kyle’s chances of getting tails on the next toss are 1 in 2.

Kyle tossed the coin 5 times and got heads each time. What were the odds of this happening?

Now we are being asked to find the probability of a both/and question, since we are being asked to identify the probability of multiple events all happening. (If it helps to picture, you can rephrase the question as: “What are the odds that BOTH his first coin tosses were heads? And What were the odds that BOTH his next tosses were heads?”, etc.)

So if we use what we know about combined probabilities, we would be able to say:

$1/2 * 1/2 * 1/2 * 1/2 * 1/2$

$1/32$

The odds are 1 in 32 (3.125%) that Kyle would have tossed heads five times in a row.

1)

2)

3)

4)

1. This is another example of an altering probability question and, again, we have two choices when it comes to solving it. Let’s go through both the algebra/proportion method and PIA.

Method 1—proportions.

We know that we must increase the number of red marbles and only red marbles, so the amount of new marbles added to the set of red marbles and to the overall total of marbles will be the same.

Our starting probability of red marbles is:

$6/18$

So now we must increase each part of our fraction by the same amount and set it equal to the desired probability of $⅗$.

${6 + x}/{18 + x} = 3/5$

$(18 + x)(3) = (6 + x)(5)$

$54 + 3x = 30 + 5x$

$24 = 2x$

$12 = x$

So we must increase our red marbles (and, consequently, the total number of marbles) by 12 in order to get a probability of $⅗$ of selecting a red marble.

To double-check this, we can plug the number back into our probability.

${6 + 12}/{18 + 12}$

$18/30$

$3/5$

We have successfully found our answer!

Our final answer is F, 12.

Method 2—PIA

The alternative method is to use plugging in answers. We will simply plug in our answer choices to increase our red marbles (and our total number of marbles) and see which answer choice results in a probability of $3/5$.

Let us start with answer choice H, 18.

${6 + 18}/{18 + 18}$

$24/36$

$2/3$

This probability is too large and any larger numbers will only get us larger probabilities. This means we can eliminate answer choices H, J, and K.

Now, let us try answer choice G, 16.

${6 + 16}/{18 + 16}$

$22/34$

$11/17$

This probability is still slightly too large. By process of elimination, our answer must be F, but let us test it to be sure.

${6 + 12}/{18 + 12}$

$18/30$

$3/5$

Success! We have found our right answer.

Our final answer is, again, F, 12.

2. Because Elliott must answer all the questions correctly, this means that this is a combination probability question. We are told that he answers each question at random, and all the questions have 3 answer options, which means that answering one question correctly has a probability of:

$1/3$

And, since this is a combination problem, answering ALL 4 questions correctly will be:

$1/3 * 1/3 * 1/3 * 1/3$

$1/81$

Our final answer is E, $1/81$

3. We have a total of 150 people and 67 of them have type A blood, while 6 of them have type AB. This means that type A blood has a probability of:

$67/150$

And type AB blood has a probability of:

$6/150$

Now we can add these probabilities together.

$67/150 + 6/150 = 73/150$

Our final answer is D, $73/150$

4. Here, we have another probability question made more complicated by the use of ratios. Again, if you need a refresher on ratios, check out our guide to ACT fractions and ratios.

First, we must find the actual number of 10th and 11th graders.

We are told that the 10th graders have a ratio of 86:255 to the school population and the 11th graders have a ratio of 18:51 to the total student population. We must first set these ratios to an equal number of total students in order to determine the number of students in each class.

We can see that the 11th graders have a reduced ratio, so we must multiply each side of the ratio by the same amount in order to equal the total number of students as the 10th graders’ ratio (255).

Luckily for us, $255/51 = 5$. This is a nice, round number to work with.

Now, we must multiply the 11th grade ratio by 5 on each side to even out the playing field.

$18(5):51(5)$

$90:255$

We are assuming for now that there are 255 students total (there may be $255 *2$ or $255 * 3$, and so forth, but this will not affect our final outcome; all that matters is that we choose a total number of students that is equal for all grades/ratios.)

So there are 86 10th graders, 90 11th graders, and the remaining students are 12th graders. Knowing that there are 255 students total, we can find the number of 12th graders by saying:

$255 - 86 - 90 = 79$

There are 79 12th graders.

This means that the probability of selecting a 10th, 11th, or 12th grader at random is:

$86/255$, $90/255$, $79/255$, respectively.

The odds are higher that the lottery will select an 11th grader, as the numerator for 11th graders is larger than that of the others.

Our final answer is B, 11th graders.