SHSAT Math Quiz: Probability Word Problems
7 questions · exam conditions
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Probability Word ProblemsQuestion 1 of 7

A bag contains 5 red marbles, 7 blue marbles, and 8 green marbles. One marble is selected at random and then replaced. A second marble is then selected. What is the probability that both marbles selected are blue?

49400\dfrac{49}{400}
720\dfrac{7}{20}
49324\dfrac{49}{324}
7100\dfrac{7}{100}
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SHSAT Math Quiz

SHSAT Math Quiz: Probability Word Problems

Practice Probability Word Problems in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Probability Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A bag contains 5 red marbles, 7 blue marbles, and 8 green marbles. One marble is selected at random and then replaced. A second marble is then selected. What is the probability that both marbles selected are blue?

  1. 49400\dfrac{49}{400} (correct answer)
  2. 720\dfrac{7}{20}
  3. 49324\dfrac{49}{324}
  4. 7100\dfrac{7}{100}
Explanation: When you encounter probability questions involving replacement, you're dealing with independent events where the outcome of the first selection doesn't affect the second. This is a key distinction from problems without replacement. To find the probability that both marbles are blue, you need to multiply the probability of selecting blue on the first draw by the probability of selecting blue on the second draw. First, determine the total number of marbles: 5 red + 7 blue + 8 green = 20 marbles total. The probability of selecting blue on the first draw is 720\frac{7}{20}. Since the marble is replaced, the probability of selecting blue on the second draw is also 720\frac{7}{20}. For both events to occur, multiply these probabilities: 720×720=49400\frac{7}{20} \times \frac{7}{20} = \frac{49}{400}, which is answer choice A. Choice B (720\frac{7}{20}) represents the probability of selecting blue on just one draw, not both. Choice C (49324\frac{49}{324}) suggests someone incorrectly calculated the total as 18 marbles instead of 20, then squared 718\frac{7}{18}. Choice D (7100\frac{7}{100}) appears to come from incorrectly treating this as an "either/or" scenario rather than "both/and." Remember: with replacement means independent events, so multiply individual probabilities. Without replacement, the sample space changes after each selection. Always verify your total count and distinguish between "and" (multiply) versus "or" (add) probability scenarios.

Question 2

In a class of 30 students, 18 play soccer, 12 play basketball, and 8 play both sports. If a student is selected at random, what is the probability that the student plays soccer given that the student plays basketball?

  1. 830\frac{8}{30}
  2. 818\frac{8}{18}
  3. 1830\frac{18}{30}
  4. 23\frac{2}{3} (correct answer)
Explanation: When you see "given that" in a probability question, you're dealing with conditional probability. This means you're finding the probability of one event happening within a restricted subset of outcomes. The question asks for P(plays soccer | plays basketball), which means "the probability a student plays soccer given that we already know they play basketball." This restricts our sample space to only the 12 basketball players, not all 30 students. Of the 12 basketball players, 8 also play soccer. Therefore, P(plays soccer | plays basketball) = 812=23\frac{8}{12} = \frac{2}{3}. This is answer choice D. Let's examine why the other choices are wrong. Choice A gives 830\frac{8}{30}, which would be the probability that a randomly selected student from the entire class plays both sports - this ignores the conditional aspect. Choice B shows 818\frac{8}{18}, which incorrectly uses the total number of soccer players as the denominator instead of focusing on the basketball players. Choice C gives 1830\frac{18}{30}, which is simply the probability that any student plays soccer, completely ignoring both the "given that" condition and the overlap between sports. Remember this key strategy: when you see "given that" in probability questions, immediately identify what group becomes your new sample space. The "given" condition tells you which subset to focus on, and you calculate the probability within that restricted group only.

Question 3

A bag contains 8 red marbles, 6 blue marbles, and 4 green marbles. If two marbles are drawn without replacement, what is the probability that the first marble is red and the second marble is blue?

  1. 851\frac{8}{51}
  2. 24153\frac{24}{153} (correct answer)
  3. 417\frac{4}{17}
  4. 1418\frac{14}{18}
Explanation: Total marbles = 18. P(first red) = 8/18. After removing one red marble, 17 marbles remain with 6 blue. P(second blue|first red) = 6/17. P(first red AND second blue) = (8/18) × (6/17) = 48/306 = 24/153. Choice A uses 17 as denominator incorrectly. Choice C simplifies incorrectly. Choice D uses addition instead of multiplication.

Question 4

Two dice are rolled simultaneously. What is the probability that the sum is greater than 8, given that both dice show the same number?

  1. 16\frac{1}{6}
  2. 26\frac{2}{6}
  3. 13\frac{1}{3} (correct answer)
  4. 1036\frac{10}{36}
Explanation: This is a conditional probability problem - you need to find the probability of one event happening given that another event has already occurred. When you see "given that" in a probability question, you're working with a restricted sample space. First, identify all outcomes where both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). That's 6 possible outcomes in your restricted sample space. Next, determine which of these outcomes have a sum greater than 8. Calculate each sum: (1,1) gives 2, (2,2) gives 4, (3,3) gives 6, (4,4) gives 8, (5,5) gives 10, and (6,6) gives 12. Only (5,5) and (6,6) have sums greater than 8 - that's 2 favorable outcomes. The probability is therefore 26=13\frac{2}{6} = \frac{1}{3}. Answer A (16\frac{1}{6}) incorrectly counts only one favorable outcome, perhaps missing either (5,5) or (6,6). Answer B (26\frac{2}{6}) gives the unreduced fraction - while mathematically equivalent to 13\frac{1}{3}, it's not the simplified form. Answer D (1036\frac{10}{36}) treats this as a regular probability problem using all 36 possible dice outcomes without applying the given condition, ignoring that we're working within the restricted space of matching dice. Remember: in conditional probability problems, always identify your new, restricted sample space first. The word "given" tells you to throw out outcomes that don't meet the condition and work only with what remains.

Question 5

A box contains 5 defective and 15 working light bulbs. If 3 bulbs are selected randomly without replacement, what is the probability that exactly 2 are defective?

  1. C(5,2)×C(15,1)C(20,3)\frac{C(5,2) \times C(15,1)}{C(20,3)} (correct answer)
  2. 5×4×1520×19×18\frac{5 \times 4 \times 15}{20 \times 19 \times 18}
  3. C(5,2)×C(15,2)C(20,3)\frac{C(5,2) \times C(15,2)}{C(20,3)}
  4. 3×C(5,2)×C(15,1)C(20,3)\frac{3 \times C(5,2) \times C(15,1)}{C(20,3)}
Explanation: Need exactly 2 defective from 5 and exactly 1 working from 15. Number of ways = C(5,2) × C(15,1). Total ways to choose 3 from 20 = C(20,3). Probability = C(5,2) × C(15,1)/C(20,3). Choice B uses ordered selection incorrectly. Choice C selects 2 working instead of 1. Choice D multiplies by unnecessary factor of 3.

Question 6

A factory produces widgets where 15% are defective. If a quality inspector randomly selects 2 widgets for testing, what is the probability that at least one is defective?

  1. 0.15+0.150.15 + 0.15
  2. 2×0.15×0.852 \times 0.15 \times 0.85
  3. 1−(0.85)21 - (0.85)^2 (correct answer)
  4. 0.1520.15^2
Explanation: When you encounter probability questions asking for "at least one" outcome, the most efficient approach is usually to use the complement rule: find the probability that the opposite occurs, then subtract from 1. Here, "at least one defective widget" means "one defective OR two defective." The complement is "zero defective widgets" (both widgets are good). Since 15% are defective, 85% are good. The probability that both randomly selected widgets are good is 0.85×0.85=(0.85)2=0.72250.85 \times 0.85 = (0.85)^2 = 0.7225. Therefore, the probability of at least one defective widget is 1−(0.85)2=1−0.7225=0.27751 - (0.85)^2 = 1 - 0.7225 = 0.2775. Choice A (0.15+0.150.15 + 0.15) incorrectly assumes you can simply add the individual probabilities. This violates probability rules because it doesn't account for the fact that these aren't mutually exclusive events. Choice B (2×0.15×0.852 \times 0.15 \times 0.85) calculates the probability of exactly one defective widget (either the first is defective and second is good, or vice versa), but this ignores the case where both widgets are defective. Choice D (0.1520.15^2) gives the probability that both widgets are defective, which is only part of what "at least one" includes. Remember: "At least one" problems on the SHSAT are almost always faster to solve using complements. Calculate the probability of "none," then subtract from 1. This avoids the tedious work of adding up multiple cases.

Question 7

A fair coin is flipped 4 times. What is the probability of getting exactly 2 heads, given that at least one head occurred?

  1. 616\frac{6}{16}
  2. 38\frac{3}{8}
  3. 14\frac{1}{4}
  4. 615\frac{6}{15} (correct answer)
Explanation: When you encounter conditional probability problems, you're looking at the probability of one event given that another event has already occurred. The key formula is: P(A|B) = P(A and B) / P(B). Here, you need P(exactly 2 heads | at least 1 head). First, find the total outcomes when flipping 4 coins: 24=162^4 = 16 possibilities. For "at least one head," it's easier to use the complement. The only way to NOT get at least one head is to get all tails (TTTT), which happens in 1 out of 16 ways. So P(at least 1 head) = 1516\frac{15}{16}, meaning 15 favorable outcomes. For "exactly 2 heads," use combinations: (42)=6\binom{4}{2} = 6 ways to choose which 2 flips are heads (HHTT, HTHT, HTTH, THHT, THTH, TTHH). Since all 6 outcomes with exactly 2 heads automatically include at least 1 head, P(exactly 2 heads AND at least 1 head) = 616\frac{6}{16}. Therefore: P(exactly 2 heads | at least 1 head) = 6/1615/16=615\frac{6/16}{15/16} = \frac{6}{15} Choice A (616\frac{6}{16}) gives the probability of exactly 2 heads without the condition. Choice B (38\frac{3}{8}) incorrectly simplifies 616\frac{6}{16}. Choice C (14\frac{1}{4}) likely confuses this with a different probability calculation. Remember: conditional probability problems require you to reduce your sample space to only the outcomes where the given condition is true, then find what portion of those satisfy your target event.