Calculate Basic Probability - TEAS Mathematics

1. Based on the information provided, what is the probability that a randomly selected student is not a nursing major and is not enrolled in Chemistry?
   Answer: $\frac{3}{10}$ This problem requires organizing the data, for example, in a two-way table. Total students = 200. Nursing majors = 120, so non-nursing majors = $200 - 120 = 80$. Enrolled in Chemistry = 70. Nursing majors in Chemistry = 50. Therefore, non-nursing majors in Chemistry = $70 - 50 = 20$. We want to find the number of students who are not nursing majors AND not in Chemistry. The total number of non-nursing majors is 80. Of these, 20 are in Chemistry, so $80 - 20 = 60$ are not in Chemistry. The number of students who are not nursing majors and not in Chemistry is 60. The probability is this number divided by the total number of students: $\frac{60}{200} = \frac{6}{20} = \frac{3}{10}$.
2. A point is chosen at random on a line segment from $x = -5$ to $x = 15$. What is the probability that the coordinate of the point, $x$, satisfies the inequality $x^2 - 4x - 12 \leq 0$?
   Answer: $\frac{2}{5}$ This geometric probability problem requires solving a quadratic inequality. First, find the total length of the line segment: $15 - (-5) = 20$ units. Next, solve the inequality $x^2 - 4x - 12 \leq 0$. Factor the quadratic: $(x - 6)(x + 2) \leq 0$. The roots are $x = 6$ and $x = -2$. Since the parabola opens upward, the inequality is satisfied between the roots, inclusive: $[-2, 6]$. The length of this favorable interval is $6 - (-2) = 8$ units. This entire interval lies within the total segment $[-5, 15]$. The probability is the ratio of the favorable length to the total length: $\frac{8}{20} = \frac{2}{5}$.
3. A jar contains 5 red marbles, 7 blue marbles, and 8 green marbles. If one marble is selected at random without looking, what is the probability the marble is blue?
   Answer: $\dfrac{7}{20}$ When you encounter probability questions involving selecting items from a collection, you need to identify the total number of possible outcomes and the number of favorable outcomes for the event you're investigating. To find the probability of selecting a blue marble, start by counting all marbles in the jar: 5 red + 7 blue + 8 green = 20 total marbles. Since you're looking for the probability of selecting blue, the number of favorable outcomes is 7 (the blue marbles). Probability equals favorable outcomes divided by total outcomes, so the probability is $\dfrac{7}{20}$. Looking at the answer choices, option A gives $\dfrac{7}{20}$, which matches our calculation perfectly. Option B shows $\dfrac{5}{20}$, which would be the probability of selecting a red marble instead of blue. Option C displays $\dfrac{8}{20}$, representing the probability of selecting a green marble. Option D presents $\dfrac{12}{20}$, which equals $\dfrac{7+5}{20}$ - this might tempt students who incorrectly add red and blue marbles together, perhaps thinking about "not green" instead of just blue. The correct answer is A: $\dfrac{7}{20}$. For TEAS probability questions, always follow this systematic approach: count the total items first, then count items matching your specific condition, and finally create your fraction. Double-check that your numerator represents exactly what the question asks for, not a related but different category.
4. A fair six-sided die is rolled one time. What is the probability of rolling a number greater than 4?
   Answer: $\dfrac{2}{6}$ When you encounter probability questions involving dice, cards, or other random events, you're working with basic probability: the number of favorable outcomes divided by the total number of possible outcomes. A standard six-sided die has faces numbered 1, 2, 3, 4, 5, and 6. To find the probability of rolling a number greater than 4, first identify which outcomes satisfy this condition. Numbers greater than 4 are 5 and 6 — that's 2 favorable outcomes. Since there are 6 total possible outcomes when rolling the die, the probability is $\dfrac{2}{6}$, which is answer choice B. Let's examine why the other options are incorrect. Choice A ($\dfrac{1}{6}$) represents the probability of rolling any specific single number, like exactly a 5 or exactly a 6, but not both together. Choice C ($\dfrac{3}{6}$) would be correct if you were looking for numbers greater than 3 (which would be 4, 5, and 6). Choice D ($\dfrac{4}{6}$) represents numbers less than or equal to 4 (which would be 1, 2, 3, and 4) — essentially the opposite of what the question asks. For TEAS probability questions, always start by clearly identifying what constitutes a "success" or favorable outcome, then count carefully. Write out the sample space if needed (1, 2, 3, 4, 5, 6) and circle the outcomes that meet your criteria. This systematic approach prevents counting errors that lead to wrong answer choices.
5. The Venn diagram shows the number of students in a class who play soccer (S), basketball (B), both or neither. If a student is chosen at random, what is the probability that the student plays basketball but NOT soccer?
   Answer: $\frac{10}{40}$ Basketball-only region = 10. Total students = 12 + 8 + 10 + 10 (neither) = 40. P = 10/40 = 1/4. Distractor B counts all basketball players (8+10=18) over total. Distractor C excludes the 'neither' region from the denominator. Distractor D makes both errors.
6. A fair six-sided die is rolled three times. What is the probability that the second roll is greater than the first roll, and the third roll is greater than the second roll?
   Answer: $\frac{5}{54}$ This problem requires careful enumeration of favorable outcomes. The total number of outcomes for three rolls is $6 \times 6 \times 6 = 216$. We need to find the number of outcomes $(a, b, c)$ where $a < b < c$, and $a, b, c$ are integers from 1 to 6. This is equivalent to choosing 3 distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$, as once the numbers are chosen, there is only one way to arrange them in increasing order. The number of ways to choose 3 distinct numbers from 6 is given by the combination formula $C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. So, there are 20 favorable outcomes. The probability is the ratio of favorable outcomes to the total number of outcomes: $\frac{20}{216}$. Simplifying the fraction by dividing the numerator and denominator by 4 gives $\frac{5}{54}$.
7. In a shipment of 120 calculators, 3 are defective. If one calculator is selected at random, what is the probability it is not defective? Express the answer as a fraction in simplest form.
   Answer: $\dfrac{39}{40}$ When you encounter probability questions, you're looking for the ratio of favorable outcomes to total possible outcomes. This question asks for the probability that a randomly selected calculator is not defective. Start by identifying what you know: 120 total calculators, 3 are defective. If 3 are defective, then 120 - 3 = 117 are not defective. The probability of selecting a non-defective calculator is therefore $\dfrac{117}{120}$. However, you need to express this in simplest form. To simplify $\dfrac{117}{120}$, find the greatest common divisor (GCD) of 117 and 120. Both numbers are divisible by 3: 117 ÷ 3 = 39 and 120 ÷ 3 = 40. Since 39 and 40 share no common factors other than 1, $\dfrac{39}{40}$ is the simplified form. Looking at the wrong answers: Choice A ($\dfrac{117}{120}$) gives the correct probability but isn't simplified—this tests whether you remember to reduce fractions. Choice C ($\dfrac{97}{100}$) might tempt you if you incorrectly converted to a percentage or made arithmetic errors. Choice D ($\dfrac{3}{120}$) represents the probability that a calculator is defective—this tests careful reading of what the question actually asks. For TEAS probability questions, always double-check three things: Are you finding the probability of the right event (defective vs. non-defective)? Did you use the correct numbers? Is your fraction in simplest form? These small details frequently distinguish correct from incorrect answers.
8. In a lottery, a player selects 4 distinct numbers from 1 to 20. The lottery then draws 4 winning numbers and 1 bonus number, all distinct, from the same range. What is the probability of a player's ticket matching exactly 3 of the 4 winning numbers and not matching the bonus number?
   Answer: $\frac{4}{323}$ This is a probability problem using combinations. First, determine the total number of ways a player can pick 4 numbers from 20: $C(20, 4) = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$. This is the sample space. For a favorable outcome, the player must choose exactly 3 of the 4 winning numbers and 1 number that is neither a winning number nor the bonus number. There are 4 winning numbers, 1 bonus number, and $20 - 4 - 1 = 15$ other numbers. The number of ways to choose 3 of the 4 winning numbers is $C(4, 3) = 4$. The number of ways to choose 1 of the 15 other numbers is $C(15, 1) = 15$. The total number of favorable outcomes is $C(4, 3) \times C(15, 1) = 4 \times 15 = 60$. The probability is $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{60}{4845}$. Simplifying by dividing numerator and denominator by 15 gives $\frac{4}{323}$.
9. A box contains colored chips as shown below. Two chips are drawn at random WITH replacement. What is the probability that both chips are the same color?
   Answer: $\frac{38}{100}$ Total = 10 chips (5 red, 3 blue, 2 green). With replacement: P(both red) = (5/10)² = 25/100; P(both blue) = (3/10)² = 9/100; P(both green) = (2/10)² = 4/100. Sum = 38/100. Distractor B adds probabilities incorrectly (5/10×3/10 etc.). Distractor C uses without-replacement mistakenly and miscalculates. Distractor D is the correct without-replacement answer, not with replacement.
10. An urn contains 6 red, 5 blue, and 4 yellow balls. If three balls are drawn without replacement, what is the probability of drawing a red, then a blue, and then a yellow ball in that specific order?
    Answer: $\frac{4}{91}$ This problem involves a sequence of dependent events. The total number of balls is $6 + 5 + 4 = 15$. The probability of the first event (drawing a red ball) is $\frac{6}{15}$. After drawing one red ball, there are 14 balls left. The probability of the second event (drawing a blue ball) is $\frac{5}{14}$. After drawing a red and a blue ball, there are 13 balls left. The probability of the third event (drawing a yellow ball) is $\frac{4}{13}$. To find the probability of this specific sequence, multiply the probabilities of each step: $P(\text{R, then B, then Y}) = \frac{6}{15} \times \frac{5}{14} \times \frac{4}{13} = \frac{2}{5} \times \frac{5}{14} \times \frac{4}{13} = \frac{2 \times 5 \times 4}{5 \times 14 \times 13} = \frac{40}{910} = \frac{4}{91}$.
11. The table shows a frequency distribution of test scores in a class. If one student's test is chosen at random, what is the probability the score is at least 80?
    Answer: $\frac{14}{40}$ At least 80' means ≥ 80, which includes 80-89 (8 students) and 90-100 (6 students). Total = 14. Total students = 5+9+12+8+6 = 40. P = 14/40. Distractor B only counts 80-89. Distractor C counts scores ≥ 70. Distractor D uses scores below 80 as denominator.
12. Four people are in a room. Assuming that a birth month is equally likely for each person, what is the probability that at least two people share the same birth month?
    Answer: $\frac{41}{96}$ This problem is best solved using the complement rule. The complement of 'at least two people share a birth month' is 'all four people have different birth months'. The total number of possible birth month assignments for the four people is $12 \times 12 \times 12 \times 12 = 12^4$. The number of ways for all four to have different birth months is a permutation: $P(12, 4) = 12 \times 11 \times 10 \times 9$. The probability that all are different is $\frac{12 \times 11 \times 10 \times 9}{12 \times 12 \times 12 \times 12} = \frac{11 \times 10 \times 9}{12 \times 12 \times 12} = \frac{990}{1728}$. Simplifying this fraction by dividing by 18 gives $\frac{55}{96}$. The probability of the original event (at least two share a birth month) is $1 - P(\text{all different}) = 1 - \frac{55}{96} = \frac{41}{96}$.
13. A standard six-sided die is rolled twice. Based on the sample space shown in the diagram, what is the probability that the sum of the two rolls is a multiple of 4?
    Answer: $\frac{1}{4}$ Sums that are multiples of 4: 4, 8, 12. Ways to get 4: (1,3),(2,2),(3,1) = 3. Ways to get 8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5. Ways to get 12: (6,6) = 1. Total = 9 out of 36 = 1/4. Distractor B miscounts by omitting one outcome (8/36). Distractor C represents 6/36. Distractor D represents 7/36 by miscounting completely.
14. A letter is chosen at random from the word "MISSISSIPPI." What is the probability the letter chosen is an "S"?
    Answer: $\dfrac{4}{11}$ When you encounter probability questions involving letters in a word, you need to count both the total number of letters and how many times your target letter appears. To find the probability of selecting an "S" from "MISSISSIPPI," first count all the letters in the word: M-I-S-S-I-S-S-I-P-P-I gives us 11 total letters. Next, count how many times "S" appears: there are 4 S's in the word. The probability is therefore $\dfrac{\text{number of S's}}{\text{total letters}} = \dfrac{4}{11}$, which is answer choice A. Let's examine why the other options are incorrect. Answer B ($\dfrac{4}{10}$) likely results from miscounting the total letters as 10 instead of 11 - perhaps by accidentally skipping one letter or thinking there are only 10. Answer C ($\dfrac{4}{9}$) suggests counting only 9 total letters, which might happen if you mistakenly exclude repeated letters or miscount significantly. Answer D ($\dfrac{4}{8}$) indicates counting just 8 total letters, which could occur if you tried to count only "unique" positions or made a major counting error. For probability questions on the TEAS, always write out or carefully count each element rather than estimating. Double-check your counting, especially with words containing repeated letters like "MISSISSIPPI." The denominator should always be the total number of items you're choosing from, while the numerator represents favorable outcomes.
15. Given that a green marble was selected, what is the probability that it came from Bag A?
    Answer: $\frac{4}{11}$ This is a conditional probability problem (related to Bayes' theorem). Let G be the event of drawing a green marble, and A be the event of choosing Bag A. We want to find $P(A|G) = \frac{P(A \text{ and } G)}{P(G)}$. First, find the overall probability of drawing a green marble, $P(G)$. $P(G) = P(G \text{ from A}) + P(G \text{ from B}) = P(G|A)P(A) + P(G|B)P(B)$. The probability of choosing either bag is $\frac{1}{2}$. $P(G|A) = \frac{4}{10}$ and $P(G|B) = \frac{7}{10}$. So, $P(G) = (\frac{4}{10})(\frac{1}{2}) + (\frac{7}{10})(\frac{1}{2}) = \frac{4}{20} + \frac{7}{20} = \frac{11}{20}$. The probability of choosing Bag A AND drawing a green marble is $P(A \text{ and } G) = P(G|A)P(A) = (\frac{4}{10})(\frac{1}{2}) = \frac{4}{20}$. Finally, $P(A|G) = \frac{4/20}{11/20} = \frac{4}{11}$.
16. A bag contains 8 red marbles and $x$ blue marbles. Two marbles are drawn from the bag without replacement. If the probability that both marbles drawn are red is $\frac{14}{33}$, what is the value of $x$?
    Answer: 4 This problem requires setting up and solving an algebraic equation based on probability principles. The total number of marbles is $8 + x$. The probability that the first marble is red is $\frac{8}{8+x}$. After drawing one red marble, there are $7+x$ marbles left, of which 7 are red. The probability that the second marble is red is $\frac{7}{7+x}$. The probability of both events occurring is $\frac{8}{8+x} \times \frac{7}{7+x} = \frac{56}{(8+x)(7+x)}$. We are given that this probability is $\frac{14}{33}$. So, $\frac{56}{(8+x)(7+x)} = \frac{14}{33}$. Divide both sides by 14: $\frac{4}{(8+x)(7+x)} = \frac{1}{33}$. Cross-multiply to get $132 = (8+x)(7+x)$. Instead of solving the quadratic $x^2 + 15x - 76 = 0$, we can test the answer choices. If $x=4$, the equation is $132 = (8+4)(7+4) = 12 \times 11 = 132$. This is correct. Thus, there are 4 blue marbles.
17. Two different integers are randomly selected from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$. What is the probability that their sum is even?
    Answer: $\frac{3}{7}$ This problem combines probability with number theory and combinations. The sum of two integers is even if both are even or if both are odd. The set contains 4 even numbers $\{2, 4, 6, 8\}$ and 4 odd numbers $\{1, 3, 5, 7\}$. First, calculate the total number of ways to choose 2 different integers from 8: $C(8, 2) = \frac{8 \times 7}{2} = 28$. Next, calculate the number of favorable outcomes. The number of ways to choose 2 even numbers is $C(4, 2) = \frac{4 \times 3}{2} = 6$. The number of ways to choose 2 odd numbers is also $C(4, 2) = 6$. The total number of favorable outcomes is the sum of these possibilities: $6 + 6 = 12$. The probability is the ratio of favorable outcomes to total outcomes: $\frac{12}{28} = \frac{3}{7}$.
18. The table shows the distribution of students in a school by grade and whether they ride the bus. If one student is selected at random from among those who ride the bus, what is the probability the student is in Grade 10?
    Answer: $\frac{45}{180}$ The sample space is restricted to bus riders (conditional probability). Total bus riders = 40+45+50+45 = 180. Grade 10 bus riders = 45. P = 45/180. Distractor B uses total students (300) as denominator. Distractor C uses Grade 10 total (75) as denominator. Distractor D confuses numerator/denominator.
19. A number is chosen at random from the integers 11 through 30 inclusive. What is the probability the number chosen is a prime?
    Answer: $\dfrac{7}{20}$ When you encounter probability questions involving prime numbers, you need to identify all prime numbers in the given range, then apply the basic probability formula: favorable outcomes divided by total outcomes. First, let's find all prime numbers from 11 through 30. A prime number has exactly two factors: 1 and itself. Testing each number: 11 (prime), 12 (divisible by 2, 3, 4, 6), 13 (prime), 14 (divisible by 2, 7), 15 (divisible by 3, 5), 16 (divisible by 2, 4, 8), 17 (prime), 18 (divisible by 2, 3, 6, 9), 19 (prime), 20 (divisible by 2, 4, 5, 10), 21 (divisible by 3, 7), 22 (divisible by 2, 11), 23 (prime), 24 (divisible by 2, 3, 4, 6, 8, 12), 25 (divisible by 5), 26 (divisible by 2, 13), 27 (divisible by 3, 9), 28 (divisible by 2, 4, 7, 14), 29 (prime), 30 (divisible by 2, 3, 5, 6, 10, 15). The primes are: 11, 13, 17, 19, 23, 29 — that's 7 prime numbers. The total integers from 11 to 30 inclusive is $30 - 11 + 1 = 20$. Therefore, the probability is $\dfrac{7}{20}$, which is answer choice C. Answer choice A ($\dfrac{5}{20}$) likely comes from missing two primes, perhaps 11 and 29. Answer choice B ($\dfrac{6}{20}$) suggests missing one prime. Answer choice D ($\dfrac{8}{20}$) indicates counting a composite number as prime, possibly including 15 or 21. Remember: when counting primes, test divisibility systematically. Numbers ending in 5 (except 5 itself) are always composite since they're divisible by 5.
20. The probability of rain on any given April day in a city is $0.3$. What is the probability it will not rain on that day?
    Answer: $0.7$ This question tests your understanding of complementary events in probability. When you encounter probability questions asking about something "not" happening, you're dealing with complementary events - two outcomes that together make up all possible outcomes. The probability of any event and its complement must always add up to $1$ (or 100%). This is because either it rains or it doesn't rain - there's no third option. So if we know the probability of rain is $0.3$, we can find the probability of no rain by subtracting from $1$: Probability of no rain = $1 - 0.3 = 0.7$ Looking at the wrong answers: Choice A ($0.3$) gives you the same probability as rain occurring, which doesn't make sense since rain and no rain can't have identical probabilities when one is $0.3$. Choice B ($0.5$) might seem reasonable if you're thinking of a 50-50 chance, but this ignores the given information that rain has a $0.3$ probability. Choice D ($1.3$) is impossible since probabilities cannot exceed $1$ - this would represent more than 100% certainty, which is mathematically impossible. The correct answer is C ($0.7$). For TEAS probability questions, remember this key formula: P(event) + P(not event) = 1. Whenever you see "not," "doesn't," or "fails to" in probability questions, you're likely dealing with complementary events. Simply subtract the given probability from $1$ to find the complement.

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