Probability
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DAT Quantitative Reasoning › Probability
A single fair die is rolled twice. What is the probability that the sum of the two rolls is at least 10?
$$\dfrac{5}{18}$$
$$\dfrac{1}{6}$$
$$\dfrac{1}{3}$$
$$\dfrac{1}{4}$$
Explanation
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.
A password consists of 3 different consonant letters followed by 2 distinct digits (0–9). All consonants and digits may be used at most once. If order matters in both parts, how many different passwords are possible?
$$,21\times20\times19\times10\times9$$
$$,\dfrac{21!}{18!}\times\dfrac{10!}{8!}$$
$$,\dfrac{21!}{18!}\times\binom{10}{2}$$
$$,21\times20\times19\times\binom{10}{2}$$
Explanation
This is a permutation problem where order matters and repetition is not allowed. When you see "order matters" and items are used "at most once," think about arranging distinct objects in sequence.
For the consonants: You need to select and arrange 3 different consonants from 21 available consonants (26 letters minus 5 vowels). Since order matters and no repetition is allowed, you have 21 choices for the first position, 20 for the second, and 19 for the third. This gives $$21 \times 20 \times 19$$, which equals $$\frac{21!}{18!}$$ (since you're taking the first 3 terms of 21 factorial).
For the digits: You need to select and arrange 2 distinct digits from 10 available digits (0-9). With order mattering and no repetition, you have 10 choices for the first digit and 9 for the second. This gives $$10 \times 9$$, which equals $$\frac{10!}{8!}$$.
The total number of passwords is the product: $$\frac{21!}{18!} \times \frac{10!}{8!}$$.
Answer A incorrectly keeps the digits as $$10 \times 9$$ instead of using factorial notation, though mathematically it's equivalent to the correct answer. Answer B uses $$\binom{10}{2}$$ for the digits, which only counts the ways to choose 2 digits but ignores their order. Answer D makes the same mistake as B by using combinations instead of permutations for the digits.
Remember: when order matters and no repetition is allowed, use permutations. The notation $$\frac{n!}{(n-r)!}$$ represents selecting and arranging r objects from n total objects.
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?
$$\frac{7}{15}$$
$$\frac{1}{4}$$
$$\frac{2}{5}$$
$$\frac{3}{8}$$
$$\frac{5}{12}$$
Explanation
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.
A dental laboratory has three quality control inspectors. Inspector X catches defects $$80%$$ of the time, Inspector Y catches defects $$75%$$ of the time, and Inspector Z catches defects $$70%$$ of the time. If a defective item is independently examined by all three inspectors, what is the probability that at least one inspector will catch the defect?
$$0.875$$
$$0.985$$
$$0.420$$
$$0.925$$
$$0.750$$
Explanation
Use the complement: P(at least one catches) = 1 - P(none catch) = 1 - (0.2)(0.25)(0.3) = 1 - 0.015 = 0.985. Choice B (0.875) comes from incorrectly using 1 - (0.2)(0.25) = 0.95 then multiplying by some factor. Choice C (0.750) results from taking the maximum individual probability. Choice D (0.925) comes from averaging the individual probabilities. Choice E (0.420) results from multiplying all the detection probabilities together.
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?
$$\frac{4}{153}$$
$$\frac{32}{765}$$
$$\frac{24}{816}$$
$$\frac{8}{204}$$
$$\frac{16}{459}$$
Explanation
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.
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?
$$\frac{2}{5}$$
$$\frac{8}{30}$$
$$\frac{3}{8}$$
$$\frac{4}{15}$$
$$\frac{8}{15}$$
Explanation
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.
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?
$$0.130$$
$$0.040$$
$$0.876$$
$$0.960$$
$$0.124$$
Explanation
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).
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?
$$\frac{4}{33}$$
$$\frac{5}{22}$$
$$\frac{1}{6}$$
$$\frac{2}{11}$$
$$\frac{3}{22}$$
Explanation
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.
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?
$$\frac{5}{8}$$
$$\frac{3}{5}$$
$$\frac{4}{5}$$
$$\frac{7}{15}$$
$$\frac{12}{25}$$
Explanation
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.
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?
$$0.9500$$
$$0.0277$$
$$0.2321$$
$$0.1353$$
$$0.7351$$
Explanation
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.