- ✓$\frac{1}{6}$
- B$\frac{5}{6}$
- C$\frac{1}{3}$
- DNone of these
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 ${}^n{C_3}.\frac{{(n - 3)\,!}}{{n\,!}} = \frac{1}{6}$.
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where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is. . . . .