Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
Each of 10 passengers board any of the three buses randomly which had no passenger initially. The probability that each bus has got at least one passenger is :
$1 - \frac{2^{40}}{3^{10}}$
$1 - \frac{^{10}C_x \times 3^7}{3^{10}}$
$\frac{P_3 3^7}{3^{10}}$
$\frac{3^{10} - 3 \cdot 2^{10} + 3}{3^{10}}$
Step-by-Step Solution
Key Concept: Apply inclusion-exclusion to count arrangements where no bus is empty by subtracting cases where buses are empty.
Total seating arrangements in buses is $3^{10}$. Using inclusion-exclusion principle, the number of favorable ways where at least one bus is non-empty is $3^{10} - 3 \cdot 2^{10} + 3$. Therefore, probability equals $\frac{3^{10} - 3 \cdot 2^{10} + 3}{3^{10}}$.
Correct Answer: 4