The Number of derangement's of set with n elements is
\(Dn = n![1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - .............+ (-1)^n \frac{1}{n!}]\)
Probability of derangement = \(\frac{Dn}{n!}\)
From the case given above the value of \(n\) = 4
So, The Probability that none of the objects occupy the place corresponding to their number
= \(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\)
= \(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\)
= \(\frac{12 - 4 + 1}{24}\) = \(\frac{9}{24}\)
= \(\frac{3}{8}\)
The correct option is (B): \(\frac{3}{8}\)
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?