Question

A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and the rest are extremely kind. A person from the group is selected at random. Find the probability that the selected person is (i) extremely patient (ii) extremely kind or honest. Which of the above you will prefer more.

Answer 1

Hint: Get the total number of persons belonging to the extremely kind category by subtracting total persons to other two categories. Probability of any event is defined mathematically as

$\text{P}=\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}$

For first part favourable cases is total number of persons who are extremely patient and for second part is sum of number of persons who are extremely kind and who are honest and hence use the above formula to get the required answer.

Complete step-by-step answer:
As there are three categories defined, a group consists of 12 persons, first one is extremely patient, second category is extremely honest and last is extremely kind. S we are given the total number of persons belong to the categories extremely patient and honest, which are
Total number of persons who are extremely patient = 3.

Total number of persons who are extremely honest = 6.

Now, we can calculate the rest number of persons belonging to extremely kind by subtracting total number of persons to extremely patient and extremely honest. So, we get
Total number of persons who are extremely kind

= 12 – (6 + 3)

= 12 – 9

= 3

Now, coming to the question:

(i) Here we need to find the probability of selecting a person at random, who is extremely patient. As we know probability can be defined mathematically as

$\text{P}=\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}...........................\left( i \right)$

Now, as we need to calculate the probability for selecting extremely patient persons out of a total number of persons. So, we know

Total number of persons who are extremely patient = 3

Total number of person in group = 12

So, favourable cases in equation (i) is 3 and total number of cases is 12. Hence, probability of selecting a extremely patient person is given as

$P=\dfrac{3}{12}=\dfrac{1}{4}$

So, answer is $\dfrac{1}{4}$

(ii) Here, we have to determine the probability to select a person who is extremely kind or honest. So, we can select any person from the category of extremely kind or extremely honest. Hence, favourable case in equation (i) is given as

Total number of persons who are extremely honest + Total number of persons who are extremely kind.

= 6 + 3

= 9

Hence, probability to select person who is extremely kind or honest is given as

$\begin{align}

  & P=\dfrac{9}{12}=\dfrac{3}{4} \\

 & P=\dfrac{3}{4} \\

\end{align}$

So, $\dfrac{3}{4}$ is the answer to this part.

Personally I will prefer to be extremely honest because if someone is honest then he or she will be patient and kind both.


Note: One may calculate the probability for the second part of the problem by adding the probabilities of extremely kind and extremely honest individually as

$\begin{align}

  & =\dfrac{6}{12}+\dfrac{3}{12}=\dfrac{9}{12} \\

 & =\dfrac{3}{4} \\

\end{align}$

Hence, it can be another approach to write the probability. One may go wrong with the above approach if someone multiplies both the individual probabilities of extremely kind and honest to get the required probability. So, be clear with the concept that if term ‘or’ is used between two or among a number of events then we need to add probabilities of individual events and if term ‘and’ is used then we need to multiply the probabilities of the events.