Probability 2 Question 16
16. The probability that, on the examination day, the student $S _1$ gets the previously allotted seat $R _1$, and
NONE of the remaining students gets the seat previously allotted to him/her is
(a) $\frac{3}{40}$
(b) $\frac{1}{8}$
(c) $\frac{7}{40}$
(d) $\frac{1}{5}$
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Solution:
- Here, five students $S _1, S _2, S _3, S _4$ and $S _5$ and five seats $R _1, R _2, R _3, R _4$ and $R _5$
$\therefore$ Total number of arrangement of sitting five students is $5 !=120$
Here, $S _1$ gets previously alloted seat $R _1$
$\therefore S _2, S _3, S _4$ and $S _5$ not get previously seats.
Total number of way $S _2, S _3, S _4$ and $S _5$ not get previously seats is
$$ \begin{aligned} 4 ! 1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !} & =241-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24} \\ & =24 \frac{12-4+1}{24}=9 \end{aligned} $$
$\therefore$ Required probability $=\frac{9}{120}=\frac{3}{40}$
