A Hospital record indicated that maternity patients stayed in the hospital for the number of days as shown in the following.

No. of days stayed	3	4	5	6	More than 6
No. of patients	15	32	56	19	5

If a patient was selected at random, find the probability that the patient stayed less than 6 days.
(a) $\dfrac{103}{127}$
(b) $\dfrac{13}{127}$
(c) $\dfrac{10}{127}$
(d) None of these.

- Hint: To begin with, we will extract the data given in the tabular form and write it in textual form. Then we will define the sample set S and find the number of elements in this sample set. Then we will define the event set A of patients staying 6 or more than 6 days and find the number of elements in event set A. Then, we will find the probability of selecting a patient who stayed for 6 or more days with the relation $\text{P}\left( \text{A} \right)=\dfrac{\text{n}\left( \text{A} \right)}{\text{n}\left( \text{S} \right)}$. Now, to find the probability of patients who stayed for less than 6 days, we will subtract P(A) from 1.

Complete step-by-step solution -
First of all, we will extract the information from the table.
From the table we can see that, the number of patients who are admitted for 3 days are 15, those admitted for 4 days are 32, 5 days are 56, 6 days are 19 and the number of patients who stayed for more than 6 days are 5.
So, the total number of patients who stayed in the hospital are (patients who stayed for 3 days + those who stayed for 4 days + those who stayed for 5 days + those who stayed for 6 days + those who stayed more than 6 days)
Therefore, the total number of patients are (15 + 32 + 56 + 19 + 5) = 127.
Let S be the event of selecting a patient at random.
Number of ways to choose a patient at random is equal to the total number of patients admitted in the hospital. This is equal to 127.
Thus, total number of elements in the event set S, given by n(S) = 127
Let A be the event of selecting a patient who stayed for 6 or more than 6 days.
Number of ways to choose such a patient will be equal to the sum of patients who stayed for 6 days and patients who stayed more than 6 days.
Therefore, the number of elements in event set A will be n(A) = (19 + 5) = 24.
Now, the probability of event A happening is given by P(A) = $\text{P}\left( \text{A} \right)=\dfrac{\text{n}\left( \text{A} \right)}{\text{n}\left( \text{S} \right)}=\dfrac{24}{127}$
Therefore, the probability of event A not happening, that is probability of selecting a patient who stays less than 6 days is given by $\text{P}\left( \text{A} \right)=1-\text{P}\left( \text{A }\!\!'\!\!\text{ } \right)$
$\Rightarrow 1-\dfrac{24}{127}=\dfrac{103}{127}$
Hence, option (a) is the correct option.

Note: Students can directly find the probability of choosing a patient who stayed less than 6 days by adding the number of patients who stayed for 3 days, those who stayed for 4 days and those who stayed for 5 days and dividing this sum by the total number of elements in the sample set.