Question
There are $$7$$ seats in a row. Three persons take seats at random. The probability that the middle seat is always occupied and no two persons are consecutive is :
A.
$$\frac{9}{{70}}$$
B.
$$\frac{9}{{35}}$$
C.
$$\frac{4}{{35}}$$
D.
none of these
Answer :
$$\frac{4}{{35}}$$
Solution :
$$\eqalign{ & n\left( S \right) = {}^7{C_3} \times 3! = \frac{{7.6.5}}{6}.6 = 210 \cr & n\left( E \right) = {}^2{C_1} \times {}^2{C_1} \times {}^1{C_1} \times 3!, \cr} $$
because one has to sit at any one of the two marked seats on the left and the other has to sit at any one of the two marked seats on the right.
$$\therefore \,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{2 \times 2 \times 6}}{{210}} = \frac{4}{{35}}.$$
$$\eqalign{ & n\left( S \right) = {}^7{C_3} \times 3! = \frac{{7.6.5}}{6}.6 = 210 \cr & n\left( E \right) = {}^2{C_1} \times {}^2{C_1} \times {}^1{C_1} \times 3!, \cr} $$
because one has to sit at any one of the two marked seats on the left and the other has to sit at any one of the two marked seats on the right.
$$\therefore \,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{2 \times 2 \times 6}}{{210}} = \frac{4}{{35}}.$$