Question

Suppose that 5 men out of 100 and 25 women out of 1000 are good orators. Assuming that there are an equal number of men and women, find the probability of choosing a good orator.

Answer 1

Hint: Here the given question is based on the concept of probability. We have to find the probability of choosing a good orator, by substituting the given data in Bayes theorem of probability and on further simplification we get the required probability of choosing a good orator.

Complete step-by-step solution:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen P}\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text {Total Number of outcomes}}}\]
Consider the given question
Let A, E1 and E2 denote the events that the person is a good orator, is a man and is a woman, respectively.
\[\,\therefore \,\,P\left( {{E_1}} \right) = \dfrac{1}{2}\] and
\[\,\,P\left( {{E_2}} \right) = \dfrac{1}{2}\]
Now, there are 5 men good orators out of 100 i.e., \[P\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{5}{{100}}\] and
there are 25 women good orators out of 1000 i.e., \[P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{{25}}{{1000}}\].
By Bayes theorem
Let \[{E_1},{\text{ }}{E_2}, \ldots ,{E_n}\;\]be a set of events associated with a sample space \[S\], where all the events \[{E_1},{\text{ }}{E_2}, \ldots ,{E_n}\;\] have nonzero probability of occurrence and they form a partition of \[S\]. Let \[A\] be any event associated with \[S\], then according to Bayes theorem,
\[P\left( {\dfrac{{{E_i}}}{A}} \right)\; = \;\dfrac{{P({E_i})P\left( {\dfrac{A}{{{E_i}}}} \right)}}{{\sum\limits_{k = 1}^n {P({E_k})P\left( {\dfrac{A}{{{E_k}}}} \right)} }}\]
Using Bayes theorem, we get
Required probability \[P\left( {\dfrac{{{E_1}}}{A}} \right)\; = \;\dfrac{{P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right)}}\]
\[ \Rightarrow \,\,\,\,\dfrac{{\dfrac{1}{2} \times \dfrac{5}{{100}}}}{{\dfrac{1}{2} \times \dfrac{5}{{100}} + \dfrac{1}{2} \times \dfrac{{25}}{{1000}}}}\]
\[ \Rightarrow \,\,\,\,\dfrac{{\dfrac{5}{{200}}}}{{\dfrac{5}{{200}} + \dfrac{{25}}{{2000}}}}\]
\[ \Rightarrow \,\,\,\,\dfrac{{\dfrac{5}{{200}}\left( 1 \right)}}{{\dfrac{5}{{200}}\left( {1 + \dfrac{5}{{10}}} \right)}}\]
On simplification, we get
\[ \Rightarrow \,\,\,\,\dfrac{1}{{1 + \dfrac{1}{2}}}\]
\[ \Rightarrow \,\,\,\,\dfrac{1}{{\dfrac{{2 + 1}}{2}}}\]
\[ \Rightarrow \,\,\,\,\dfrac{2}{3}\]

Hence, the required probability of choosing a good orator is \[\dfrac{2}{3}\].

Note: The probability is a number of possible values. Candidate must know we have to use the permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem. Here we arrange the things in the possible ways so we are using the concept permutation.