Question

An elevator starts with 4 passengers and stops at 4 floors. Find the probability of the following events:

1) all passengers leave at the same floor (Event A)

2) all passengers leave at the two floors (Event B)

3) all passengers leave at the three floors (Event C)

4) all passengers leave at different floors (Event D)

Answer 1

$1:P\left( A \right) =\left[ \begin{array}{c} 4 possible\,\,floors\\ 4 passengers\\ total\,\,variants\,\,4^4\\\end{array} \right] =\frac{4}{4^4}=0.015625\\2:P\left( B \right) =\left[ \begin{array}{c} C_{4}^{2}=6\,\,variants\,\,of\,\,floors\\ 2^4-2=14\left( for\,\,each\,\,one\,\,floor\,\,but\,\,all\,\,on\,\,the\,\,1st\,\,or\,\,all\,\,on\,\,the\,\,2nd\,\,is\,\,not\,\,ok \right)\\\end{array} \right] =\frac{6\cdot 14}{4^4}=0.328125\\4:P\left( D \right) =\left[ 4!=24\,\,variants \right] =\frac{24}{4^4}=0.09375\\3:P\left( C \right) =1-P\left( A \right) -P\left( B \right) -P\left( D \right) =1-0.015625-0.328125-0.09375=0.5625$