Probability - DAT Quantitative Reasoning

1. A bag contains $8$ red marbles, $6$ blue marbles, and $4$ green marbles. Three marbles are drawn consecutively without replacement. What is the probability that the first marble is red, the second is blue, and the third is green?
   Answer: $\frac{4}{153}$ Total marbles = 18. P(Red first, Blue second, Green third) = (8/18) × (6/17) × (4/16) = 192/4896 = 4/153. Choice B results from incorrect denominator calculation. Choice C comes from using wrong total counts. Choice D results from not reducing the fraction properly. Choice E comes from computational errors in the multiplication.
2. Refer to the deck configuration shown. From a specially prepared 20-card deck (shown below), two cards are drawn without replacement. What is the probability that exactly one of the two cards is a 'Star' card?
   Answer: $\frac{48}{95}$ There are 8 Star cards and 12 Non-Star cards. For exactly one Star: P(Star then Non-Star) + P(Non-Star then Star) = (8/20)(12/19) + (12/20)(8/19) = 96/380 + 96/380 = 192/380 = 48/95. Alternatively, using combinations: C(8,1)×C(12,1)/C(20,2) = (8×12)/190 = 96/190 = 48/95.
3. In a clinical trial, $60\%$ of patients respond to Treatment A, $40\%$ respond to Treatment B, and $25\%$ respond to both. If a patient responds to at least one treatment, what is the probability that the patient responds to Treatment A?
   Answer: $\frac{4}{5}$ P(responds to at least one) = 0.6 + 0.4 - 0.25 = 0.75. P(A|at least one) = P(A)/P(at least one) = 0.6/0.75 = 4/5. Choice B (3/5) is P(A) without conditioning. Choice C (12/25) results from incorrect calculation. Choice D (7/15) comes from using wrong probability values. Choice E (5/8) results from computational errors.
4. Use the Venn diagram shown. A student is chosen at random from a class of 50. What is the probability that the student takes Chemistry but NOT Biology, given that the student takes at least one of the two subjects?
   Answer: $\frac{12}{42}$ From the Venn diagram: Biology only = 18, Chemistry only = 12, Both = 12, Neither = 8. Total taking at least one = 18 + 12 + 12 = 42. P(Chem only | at least one) = 12/42. (B) uses total class instead of conditioning. (C) uses Biology only. (D) uses 30 (total Chemistry) as denominator.
5. A medical device has a $2\%$ failure rate. If $10$ such devices are used independently, what is the probability that exactly $2$ devices fail?
   Answer: $0.0153$ P(exactly 2 fail) = C(10,2) × (0.02)² × (0.98)⁸ = 45 × 0.0004 × 0.8508 = 0.0153. Choice B (0.0036) results from not including the binomial coefficient. Choice C (0.2000) is 10 × 0.02. Choice D (0.8171) is approximately (0.98)¹⁰. Choice E (0.1667) results from computational errors.
6. A single fair die is rolled twice. What is the probability that the sum of the two rolls is at least 10?
   Answer: $\dfrac{1}{6}$ When you encounter probability questions involving dice rolls, you need to systematically count favorable outcomes and divide by total possible outcomes. With two dice rolls, there are $6 \times 6 = 36$ total possible outcomes. To find the probability of getting a sum of at least 10, identify all ways to achieve sums of 10, 11, or 12. For sum = 10: (4,6), (5,5), (6,4) — that's 3 outcomes. For sum = 11: (5,6), (6,5) — that's 2 outcomes. For sum = 12: (6,6) — that's 1 outcome. Total favorable outcomes: 3 + 2 + 1 = 6. Therefore, the probability is $\frac{6}{36} = \frac{1}{6}$, making choice A correct. Choice B ($\frac{1}{4}$) equals $\frac{9}{36}$, which would suggest 9 favorable outcomes — this likely comes from incorrectly counting or including impossible combinations. Choice C ($\frac{5}{18}$) equals $\frac{10}{36}$, suggesting 10 favorable outcomes — this might result from double-counting some combinations or including sums of 9. Choice D ($\frac{1}{3}$) equals $\frac{12}{36}$, indicating 12 favorable outcomes — this error often occurs when students mistakenly include all sums from 9 to 12 instead of just 10 to 12. Remember to list outcomes systematically when dealing with "at least" or "at most" probability questions. Draw a table if needed, and always double-check that your favorable outcomes actually satisfy the given condition. The key is being methodical rather than rushing through the counting.
7. A box contains $12$ dental tools: $5$ extractors, $4$ scalers, and $3$ explorers. If $3$ tools are selected randomly without replacement, what is the probability that exactly $2$ are extractors and $1$ is a scaler?
   Answer: $\frac{2}{11}$ P(2 extractors, 1 scaler) = [C(5,2) × C(4,1) × C(3,0)]/C(12,3) = (10 × 4 × 1)/220 = 40/220 = 2/11. Choice B (5/22) results from computational errors in combinations. Choice C (3/22) comes from using wrong combination values. Choice D (4/33) results from incorrect total combinations. Choice E (1/6) comes from oversimplified probability calculation.
8. Two dice are rolled simultaneously. Given that the sum is greater than $8$, what is the probability that both dice show the same number?
   Answer: $\frac{2}{10}$ Sums > 8: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) - total 10 outcomes. Same numbers with sum > 8: (5,5), (6,6) - total 2 outcomes. P(same|sum > 8) = 2/10 = 1/5. Choice B (1/6) is P(same) unconditionally. Choice C (3/10) results from counting errors. Choice D (5/18) comes from incorrect calculations. Choice E (1/3) results from oversimplified reasoning.
9. A dental practice schedules $5$ appointments per day. Each patient has a $15\%$ probability of canceling independently. What is the probability that exactly $1$ patient cancels?
   Answer: $0.3915$ P(exactly 1 cancels) = C(5,1) × (0.15)¹ × (0.85)⁴ = 5 × 0.15 × 0.5220 = 0.3915. Choice B (0.7500) comes from 5 × 0.15. Choice C (0.1500) is the individual cancellation rate. Choice D (0.4437) is approximately (0.85)⁴. Choice E (0.5220) results from computational errors.
10. A fair coin is tossed until a head appears or until 3 tosses are made. What is the expected number of tosses?
    Answer: 1.75 With probability 1/2, the first toss is heads and X=1. With probability 1/4, you get tails then heads and X=2. Otherwise, the first two tosses are tails, so the third toss happens regardless and X=3. Expected tosses = 1(1/2) + 2(1/4) + 3(1/4) = 0.5 + 0.5 + 0.75 = 1.75. The tempting 2.00 wrongly treats the process as continuing until a head with no three-toss limit.
11. A dental clinic has a $95\%$ success rate for root canals. If $6$ independent root canal procedures are performed, what is the probability that exactly $5$ are successful?
    Answer: $0.2321$ P(exactly 5 successful) = C(6,5) × (0.95)⁵ × (0.05)¹ = 6 × 0.7738 × 0.05 = 0.2321. Choice B (0.7351) results from calculating (0.95)⁵. Choice C (0.9500) is the individual success rate. Choice D (0.0277) comes from computational errors. Choice E (0.1353) results from using wrong binomial coefficient.
12. In a dental clinic, the probability that a patient needs a filling is $0.3$, needs a crown is $0.15$, and needs both a filling and a crown is $0.08$. Given that a randomly selected patient needs a crown, what is the probability that this patient also needs a filling?
    Answer: $\frac{8}{15}$ This is conditional probability: P(filling|crown) = P(filling and crown)/P(crown) = 0.08/0.15 = 8/15. Choice B (4/15) results from incorrectly using P(crown)/P(filling). Choice C (2/5) comes from using P(filling) directly without conditioning. Choice D (8/30) results from not simplifying the fraction. Choice E (3/8) comes from reversing the conditional probability calculation.
13. A dental student must pass $4$ independent practical exams, each with a $0.8$ probability of passing. What is the probability that the student passes exactly $3$ out of the $4$ exams?
    Answer: $0.4096$ This is a binomial probability: P(exactly 3 passes) = C(4,3) × (0.8)³ × (0.2)¹ = 4 × 0.512 × 0.2 = 0.4096. Choice B (0.2048) results from using C(4,3) = 2 instead of 4. Choice C (0.8192) comes from calculating P(at least 3). Choice D (0.1536) results from computational errors. Choice E (0.0256) comes from calculating (0.8)³ × (0.2) without the binomial coefficient.
14. The table below shows results of testing 200 patients for a disease. What is the probability that a patient actually has the disease given that they tested positive?
    Answer: $\frac{18}{33}$ From the table: 18 tested positive and have the disease; 15 tested positive but don't have the disease. Total positive tests = 33. P(disease | positive) = 18/33. (B) is sensitivity P(positive | disease). (C) is joint probability. (D) is false positive rate among positives.
15. Two dental laboratories independently produce crowns with defect rates of $5\%$ and $8\%$ respectively. If one crown is randomly selected from each laboratory, what is the probability that at least one crown is defective?
    Answer: $0.124$ P(at least one defective) = 1 - P(both good) = 1 - (0.95)(0.92) = 1 - 0.874 = 0.126 ≈ 0.124. Choice B (0.040) results from multiplying the defect rates. Choice C (0.130) comes from adding the defect rates. Choice D (0.876) is P(both good). Choice E (0.960) results from 1 - (0.05 × 0.08).
16. A box contains $10$ identical balls except for color: $4$ are red, $3$ are blue, and $3$ are green. Four balls are selected simultaneously. What is the probability that exactly $2$ are red and exactly $1$ is blue?
    Answer: $\frac{9}{35}$ Total ways to select 4 balls from 10: C(10,4) = 210. Ways to select exactly 2 red, 1 blue, 1 green: C(4,2)×C(3,1)×C(3,1) = 6×3×3 = 54. P(2 red, 1 blue, 1 green) = 54/210 = 9/35. Choice A (18/70) results from computational errors. Choice B (36/210) is the unreduced fraction with wrong numerator. Choice C (12/70) comes from incorrect combination calculation. Choice D (6/35) results from using wrong total combinations.
17. A bag has 3 red, 4 blue, and 5 green marbles. Drawing 3 without replacement, what is P(all different colors)?
    Answer: $\frac{3}{11}$ There are 12 marbles total. Favorable draws are one red, one blue, and one green: 3 times 4 times 5 = 60 ways. Total ways to draw 3 marbles is 12 choose 3 = 220, so 60/220 = 3/11. The tempting wrong answer 1/22 comes from multiplying 3/12 times 4/11 times 5/10 for one fixed color order, but the colors can come in any of 6 orders, so multiply by 6.
18. The probability distribution table for discrete random variable X is shown. What is $P(X \geq 3 \mid X \geq 2)$?
    Answer: $\frac{0.50}{0.75}$ From the table: $P(X \geq 2) = 0.25+0.30+0.20 = 0.75$. $P(X \geq 3) = 0.30+0.20 = 0.50$. Conditional probability = $0.50/0.75 = 2/3$. (B) forgets conditioning. (C) uses $P(X=3)$ only. (D) miscomputes both.
19. A medical research study involves testing a new dental treatment on patients. The probability that a patient responds positively to Treatment A is $0.6$, to Treatment B is $0.4$, and to both treatments is $0.25$. If a patient is selected at random and responds positively to Treatment A, what is the probability that this patient also responds positively to Treatment B?
    Answer: $\frac{5}{12}$ This is a conditional probability problem. We need P(B|A) = P(A and B)/P(A) = 0.25/0.6 = 25/60 = 5/12. Choice B (1/4) results from using P(A and B) directly. Choice C (3/8) comes from incorrectly calculating (0.6 × 0.4)/0.64. Choice D (2/5) results from using P(B)/P(A). Choice E (7/15) comes from adding probabilities incorrectly.
20. Refer to the grid shown below. A particle starts at point A (lower-left) and moves to point B (upper-right) by taking unit steps only right or up. If each valid path is equally likely, what is the probability that the particle passes through point C?
    Answer: $\frac{18}{35}$ A to B requires 4 rights and 3 ups: $\binom{7}{3}=35$ paths. A to C (2 right, 1 up): $\binom{3}{1}=3$ paths. C to B (2 right, 2 up): $\binom{4}{2}=6$ paths. Paths through C: $3 \times 6 = 18$. Probability = 18/35. (B) assumes symmetry. (C) forgets to multiply by C-to-B paths. (D) uses wrong binomial coefficient.

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