Statistics And Probability Question 19
Question: n letters to each of which corresponds on addressed envelope are placed in the envelop at random. Then the probability that n letter is placed in the right envelope, will be:
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Probability of n balls $ =1-P(A_1EA_2EA_3E…EA_{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-[ \frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}…(-1)\frac{n-1!}{n!} ] $ $ =\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-….+{{(-1)}^{n}}\frac{1}{n!} $
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