A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?

### Concept & Probability of the Complement If a question asks for the probability that "none" are of a specific type, calculate the probability of drawing from the subset that excludes that type. $$P(\text{None are } X) = \frac{^{\text{Total Non-}X}C_r}{^{\text{Total Items}}C_r}$$ ### Step-by-Step Solution 1. **Total Items:** The basket has $6$ blue, $2$ red, $4$ green, and $3$ yellow balls. Total $= 6 + 2 + 4 + 3 = 15$ balls. 2. **Total Outcomes:** Picking $3$ balls from $15$. Total ways $= ^{15}C_3$. $$^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$$ 3. **Favorable Outcomes:** We want no yellow balls. Total non-yellow balls $= 15 - 3 = 12$ balls. Number of ways to pick $3$ balls from these $12$ non-yellow balls $= ^{12}C_3$. $$^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$$ 4. **Probability:** $$P(\text{none is yellow}) = \frac{220}{455}$$ 5. **Simplify:** Divide numerator and denominator by $5$. $$P = \frac{44}{91}$$ ### Exam Strategy & Shortcut To quickly simplify $\frac{220}{455}$, note that both end in $0$ or $5$, making them divisible by $5$. $220/5 = 44$. Since $44$ is $4 \times 11$ and $455/5 = 91$ ($7 \times 13$), there are no further common factors. ### Common Pitfall Trying to calculate the probability of picking exactly 1, 2, or 3 yellow balls and subtracting from 1 is valid but highly inefficient. Always take the direct approach when calculating "none." ### Final Answer Therefore, the correct answer is **$\frac{44}{91}$**.