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Example Questions

Example Question #1548 : Psat Mathematics

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

Possible Answers:

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{3}{14}$

$\frac{5}{14}$

$\frac{2}{14}$

Correct answer:

$\frac{5}{14}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

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Example Question #52 : Calculating Discrete Probability

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?

Possible Answers:

$\frac{5}{13}$

$\frac{1}{7}$

$\frac{5}{14}$

$\frac{6}{7}$

$\frac{17}{14}$

Correct answer:

$\frac{6}{7}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)$

The interesection is:

$P(A\bigcap B)=P(A)P(B)=\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

So, we can find the probability of drawing at least one consonant:

$P(A\bigcup B)= \frac{5}{7}+\frac{1}{2}-\frac{5}{14}=\frac{10}{14}+\frac{7}{14}-\frac{5}{14}=\frac{6}{7}$

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Example Question #1851 : Problem Solving Questions

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is drawn from Set A, and one from Set B. What is the probability of drawing a matching pair of letters?

Possible Answers:

$\frac{4}{13}$

$\frac{2}{13}$

$\frac{1}{7}$

$\frac{1}{21}$

$\frac{1}{42}$

Correct answer:

$\frac{1}{21}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, 7(6)=42 .

Therefore, the probability of drawing a matching set is:

$\frac{2}{42}=\frac{1}{21}$

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Example Question #51 : Probability

In a particular high school, 200 students are freshmen, 150 students are sophomores, 250 students are juniors, and 100 students are seniors. Twenty percent of freshmen are in honors classes, ten percent of sophomores are in honors classes, twelve percent of juniors are in honors classes, and thirty percent of seniors are in honors classes.

If a student is chosen at random, what is the probability that that student will be a student who attends honors classes?

Possible Answers:

$\frac{23}{70}$

$\frac{9}{100}$

$\frac{23}{140}$

$\frac{9}{50}$

$\frac{5}{7}$

Correct answer:

$\frac{23}{140}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

$\frac{200(.20)+(150)(.10)+(250)(.12)+(100)(0.30)}{700}$

$\frac{115}{700}=\frac{23}{140}$

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Example Question #61 : Discrete Probability

If a student is chosen at random, what is the probability that that student will be a senior student and a student who does not attend honors classes?

Possible Answers:

$\frac{3}{10}$

$\frac{4}{35}$

$\frac{2}{5}$

$\frac{3}{70}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The percentage of seniors that do not attend honors classes is:

(1-.30)=.70

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

$\frac{100}{700}(.70)=\frac{1}{10}$

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Example Question #161 : Data Analysis

There are 4 blue marbles, 5 green marbles, 7 black marbles, and 3 white marbles in a jar. If you reach in and grab out one marble, what is the probabilty of drawing a green marble?

Possible Answers:

$\frac{4}{19}$

$\frac{5}{19}$

$\frac{5}{16}$

$\frac{1}{5}$

$\frac{14}{19}$

Correct answer:

$\frac{5}{19}$

Explanation:

To calculate the probability of an event happening, take the number of ways to draw a green marble (5, because there are 5 possible green marbles to choose) by the total number of ways to draw a marble from the jar (19). Thus the answer is $\frac{5}{19}$

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Example Question #162 : Data Analysis

A fair, six sided die is number with the numbers one through three with each number appearing twice. What is the chance that you roll a two when rolling the die one time?

Possible Answers:

$\frac{2}{3}$

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

To calculate a probability, divide the possible ways to get the desire outcome by the total possible number of outcomes. We only care about rolling a 2, which can happen in 2 different ways (because there are two 2s on the die).

Thus we get:

$\frac{2}{6}=\frac{2\cdot 1}{2\cdot 3}=\frac{1}{3}$

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Example Question #53 : Probability

When rolling $\textup{2 fair dice}$ , what is the probability that the sum of their faces will be ? Reduce all fractions.

Possible Answers:

$\frac{1}{6}$

$\frac{1}{7}$

$\frac{1}{36}$

$\frac{7}{6}$

$\frac{7}{36}$

Correct answer:

$\frac{1}{6}$

Explanation:

To find the probability that two dice sum to seven we need to figure out the total number of ways that can happen. Represent the rolls as an ordered pair, with the first number in the pair corresponding to the first roll, and the second to the second roll. Then all the ways to get a sum of 7 are as follows:

(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) - or 6 different ways.

There are 6*6 = 36 possible outcomes when rolling two dice (6 different ways the first roll could come out, and 6 ways the second roll could come out)
thus the probability the sum is 7 is:

$\frac{6}{36}=\frac{1}{6}$

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Example Question #53 : How To Find The Probability Of An Outcome

There are $14\textup{ cups}$ , $\textup{33 dishes}$ , and $\textup{41 spoons}$ in a box. What is the chance of someone drawing out a dish then a spoon, presuming that he or she does not place items back into the box after drawing them out? Round to the nearest hundredth of a percent.

Possible Answers:

$\small \small 47.13\%$

$\small 17.67\%$

$\small 18.04\%$

$\small 17.47\%$

$\small 16.33\%$

Correct answer:

$\small 17.67\%$

Explanation:

For a probability question like this, first do a sum of the total possible outcomes. For the data given, this is $\small 14+33+41$ or $\small 88$ . Now, the probability of drawing a dish is $\small \frac{33}{88}$ . After this, there are $\small 87$ items in the box. The probability of drawing a spoon after this is $\small \small \frac{41}{87}$ .

Now, when two events are sequential like this, the probability of the two together is calculated by multiplying their respective probabilities. Thus, your total probability is $\small \small \frac{33}{88} * \frac{41}{87} = \frac{1353}{7656}$ or $\small 17.67\%$

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Example Question #51 : How To Find The Probability Of An Outcome

There are $\textup{35 tomatoes}$ , $\textup{55 onions}$ , and $\textup{33 pears}$ on a shelf. What is the probability of picking a tomato at random? Round to the nearest hundredth of a percent.

Possible Answers:

$\small 59.34\%$

$\small 21.94\%$

$\small 15.32\%$

$\small 63.64\%$

$\small 28.46\%$

Correct answer:

$\small 28.46\%$

Explanation:

For a probability question like this, first do a sum of the total possible outcomes. For the data given, this is $\small 35+55+33$ or $\small 123$ . Now, the probability of drawing a tomato is $\small \small \frac{35}{123}$ . Therefore, it is $\small 28.46\%$ .

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ACT Math : ACT Math

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1548 : Psat Mathematics

Example Question #52 : Calculating Discrete Probability

Example Question #1851 : Problem Solving Questions

Example Question #51 : Probability

Example Question #61 : Discrete Probability

Example Question #161 : Data Analysis

Example Question #162 : Data Analysis

Example Question #53 : Probability

Example Question #53 : How To Find The Probability Of An Outcome

Example Question #51 : How To Find The Probability Of An Outcome

All ACT Math Resources

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