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14.Probability
hard
There are $n$ different objects $1, 2, 3,......n$ distributed at random in $n$ places marked $1, 2, 3, ......n$. The probability that at least three of the objects occupy places corresponding to their number is
A
$\frac{1}{6}$
B
$\frac{5}{6}$
C
$\frac{1}{3}$
D
None of these
Solution
(a) Let ${E_i}$ denote the event that the ${i^{th}}$ object goes to the ${i^{th}}$ place,
we have $P({E_i}) = \frac{{(n – 1)\,!}}{{n\,!}} = \frac{1}{n},\forall \,\,i$
and $P({E_1} \cap {E_j} \cap {E_l}) = \frac{{(n – 3)\,\,!}}{{n\,\,!}}$ for $i < j < k$
Since we can choose $3$ places out of $n$ in ${}^n{C_3}$ ways.
The probability of the required event is ${}^n{C_3}.\frac{{(n – 3)\,!}}{{n\,!}} = \frac{1}{6}$.
Standard 11
Mathematics
