A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?

### Concept & The "At Least One" Rule Finding the probability of "at least one" condition is best solved using the complement approach. We find the probability that the condition is completely absent and subtract it from 1. $$P(\text{at least one}) = 1 - P(\text{none})$$ ### Step-by-Step Solution 1. **Total Items:** $6$ blue, $2$ red, $4$ green, $3$ yellow. Total $= 15$ balls. 2. **Total Outcomes:** Drawing $5$ balls from $15$. $$^{15}C_5 = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$$ 3. **Favorable Outcomes for Complement (No blue):** Total non-blue balls $= 15 - 6 (\text{blue}) = 9$ balls. - Ways to pick $5$ non-blue balls $= ^9C_5$. By symmetry, $^9C_5 = ^9C_4$. $$^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$ 4. **Probability of Complement:** $$P(\text{no blue}) = \frac{126}{3003}$$ - Divide by $3$: $\frac{42}{1001}$ - Divide by $7$: $\frac{6}{143}$ 5. **Final Probability:** $$P(\text{at least one blue}) = 1 - \frac{6}{143} = \frac{143 - 6}{143} = \frac{137}{143}$$ ### Exam Strategy & Shortcut The property $^nC_r = ^nC_{n-r}$ is incredibly useful here. Calculating $^9C_4$ instead of $^9C_5$ saves you from multiplying and dividing an extra term. Additionally, recognizing $1001 = 7 \times 11 \times 13$ makes simplifying fractions much faster. ### Common Pitfall Failing to simplify the fraction before subtracting from $1$. Subtracting $\frac{126}{3003}$ from $1$ yields $\frac{2877}{3003}$, which is much harder to recognize as the correct multiple-choice option without extensive simplification. ### Final Answer Therefore, the correct answer is **$\frac{137}{143}$**.