A) \[\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+...{{(-1)}^{n}}\frac{1}{n!}\]
B) \[\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+...\frac{1}{n!}\]
C) \[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+...{{(-1)}^{n}}\frac{1}{n!}\]
D) None of these
Correct Answer: C
Solution :
[c] Probability of n balls \[=1-P({{A}_{1}}E{{A}_{2}}E{{A}_{3}}E...E{{A}_{n}})\] Where \[{{A}_{1}}...{{A}_{n}}\] the event that the letter is placed at right envelope. \[=1-[\Sigma P({{A}_{i}})-\Sigma P({{A}_{i}}\cap {{A}_{k}})\] \[+\Sigma P({{A}_{i}}\cap {{A}_{j}}\cap {{A}_{k}})...+{{(-1)}^{n-1}}P({{A}_{i}}\cap {{A}_{j}}\cap {{A}_{n}})]\] Here, \[P({{A}_{i}})=\frac{(n-1)!}{n!}\] \[P({{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}}\cap ...\cap {{A}_{n}})=\frac{(n-r)!}{n!}\] \[\Rightarrow \Sigma \overline{{{A}_{1}}}\cap \overline{{{A}_{2}}}\cap \overline{{{A}_{3}}}\cap ....\cap \overline{{{A}_{n}}}\] \[=1-\left[ \frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}...(-1)\frac{n-1!}{n!} \right]\] \[=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-....+{{(-1)}^{n}}\frac{1}{n!}\]
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