Question

1. Suppose that ten passengers are boarding seven buses at random (a) Treating the passe ngers as indistinguishable, find the probability that there is at least one passenger in each bus (b) Treating the passengers as distinguishable, find the probability that there is at least one passenger in Hint: Use the inclusion-exclusion rule. buses number 1, 2 and 3

Answer 1

a) here let x1,x2..x7 pessengers are on 7 buses

therefore x1+x2+x3...+x7 =10

number of positive solution of above =$\binom{10-1}{7-1}$ =$\binom{9}{6}$=84

b)

total number of ways to accommodate 10 passengers in 7 buses =7¹⁰ =282475249(as each passenger has 7 choice)

number of ways so that there is no passenger in bus 1,2,3 =4¹⁰ =1048576

hence number of ways at least one passenger in bus number 1,2 and 3 =282475249-1048576=281426673