A minibus has 9 passenger seats. The probability of a seat being occupied is estimated to be 0.63. Calculate the probability that on a typical run:
P(0 seats being occupied) $$ {}_9C_0*.63^0*(1-.63)9^{-0} $$ This give me the answer to no seats being taken on the bus as $1.29962...*10^{-4}$
which is way off but I do not see where I am calculating wrong?
You should be computing $$ \mathbb{P}[N=0] = \binom{9}{0} 0.63^0 (1-0.63)^9 = 0.37^9 \approx 0.00012996174 $$
Why do you think this is way off?