Question

There are n different objects 1 , 2 , 3 , ……, n distributed at random in n places marked 1 , 2 , 3 ,….., n . If p is the probability that at least three of the objects occupy places corresponding to their number, find 6p.

Answer 1

Assuming $E_i$ be the event that the ith object goes to the ith place , we get $P\left( {{E_i} \cap {E_j} \cap {E_k}} \right) = \dfrac{{\left( {n - 3} \right)!}}{{n!}}{\text{ }}for{\text{ }}i < j < k$ and in order to find the probability that at least three objects occupy places corresponding to their number , as 3 places out of n places can be chosen in ${}^n{C_3}$ ways the probability is given by $p = \dfrac{{\left( {n - 3} \right)!}}{{n!}}{}^n{C_3}$ and multiplying this by 6 we get the required value .

Complete step by step solution:
We are given that n different objects 1 , 2 , 3 , ……, n distributed at random in n places marked 1 , 2 , 3 ,….., n
So let ${E_i}$ be the event that the ith object goes to the ith place
Hence , we need to find the probability that at least three objects occupy places corresponding to their number
That is the events ${E_i},{E_j},{E_k}$ occurs at the same time
This can be given by $P\left( {{E_i} \cap {E_j} \cap {E_k}} \right) = \dfrac{{\left( {n - 3} \right)!}}{{n!}}{\text{ }}for{\text{ }}i < j < k$
We know that we can choose 3 places out of n places in ${}^n{C_3}$ ways
Therefore the probability p is given by
$\Rightarrow p = \dfrac{{\left( {n - 3} \right)!}}{{n!}}{}^n{C_3}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Substituting we get,
$\Rightarrow p = \dfrac{{\left( {n - 3} \right)!}}{{n!}}\dfrac{{n!}}{{\left( {n - 3} \right)!3!}} \\\ \Rightarrow p = \dfrac{1}{{3!}} \\\ \Rightarrow p = \dfrac{1}{{3 \times 2 \times 1}} = \dfrac{1}{6} \\\$
Hence we get the value of p.
We are asked for the value of 6p.
$\Rightarrow 6p = 6\left( {\dfrac{1}{6}} \right) = 1$

Hence our required value is 1.

Note:
Conditional probability is the probability of one event occurring with some relationship to one or more other events.
Permutations are for lists (order matters) and combinations are for groups (order doesn't matter).