Solution: The probability of selecting no quantum dots (all nanotubes) is $\frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220}$. The complement gives the probability of at least one quantum dot: $1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}$. \boxed{\dfrac{37}{44}}

February 22, 2026

$Solution: The probability of selecting no quantum dots (all nanotubes) is $\frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220}$. The complement gives the probability of at least one quantum dot: $1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}$. \boxed{\dfrac{37}{44}}$

#### 50Question: A software compliance consultant must verify 12 data modules, 3 of which are non-compliant. If 4 modules are randomly selected for audit, what is the probability that exactly 2 are non-compliant?
Solution: The total number of ways to choose 4 modules from 12 is $\binom{12}{4}$. The number of favorable outcomes (2 non-compliant and 2 compliant) is $\binom{3}{2} \cdot \binom{9}{2}$. Thus, the probability is $\frac{\binom{3}{2} \cdot \binom{9}{2}}{\binom{12}{4}} = \frac{3 \cdot 36}{495} = \frac{108}{495} = \frac{12}{55}$. \boxed{\dfrac{12}{55}}
Question: A nanotechnologist is designing a material with 5 types of quantum dots and 7 types of nanotubes. If 3 components are randomly selected, what is the probability that at least one is a quantum dot?
Question: A venture capitalist evaluates 9 clean energy startups, 4 of which are solar-based. If 5 startups are chosen randomly, what is the probability that exactly 3 are solar-based?
Solution: The total number of ways to select 5 startups is $\binom{9}{5}$. The favorable case is $\binom{4}{3} \cdot \binom{5}{2}$ (3 solar and 2 non-solar). Thus, the probability is $\frac{4 \cdot 10}{126} = \frac{40}{126} = \frac{20}{63}$. \boxed{\dfrac{20}{63}}### Question 1
A rectangular garden has a length that is twice its width. If the width is 10 meters, what is the area of the garden?

Solution: The probability of selecting no quantum dots (all nanotubes) is $\frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220}$. The complement gives the probability of at least one quantum dot: $1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}$. \boxed{\dfrac{37}{44}}

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