Question

In a group of 20 people, there are 5 graduates. If 3 people are selected at random from the group, the probability that at least one is graduated is?
(a) $\dfrac{115}{228}$
(b) $\dfrac{135}{228}$
(c) $\dfrac{225}{228}$
(d) $\dfrac{137}{228}$

Answer 1

Hint: Consider three cases that are possible for the given condition. In the first case assume that one person has graduated and the other two have not graduated. Use the formula of combinations ${}^{n}{{C}_{r}}$ to select the people and find the probability by dividing it with ${}^{20}{{C}_{3}}$ which is the total number of ways of selecting 3 people from 20 people. Similarly, form the other two cases in which two people are graduated and all the three people are graduated respectively, find their probabilities. Take the sum of the probabilities in all the three cases to get the answer.

Complete step by step answer:
Here we have been provided with 20 people with 5 graduated people. We are asked to find the probability of selecting at least one graduated person if we select 3 people at random.
Now, at least one person is graduated means it may be possible that two are graduated or all the three people selected are graduated. Let us see the following cases possible.
(1) Here we assume that one person selected turns out to be graduated, which means the other two selected are not graduated. Since there are 5 graduates so 15 are not graduates, therefore using the formula of combinations we have,
$\Rightarrow $ Number of ways to select 1 graduate from 5 graduates and 2 non graduates from 15 non graduates simultaneously = ${}^{5}{{C}_{1}}\times {}^{15}{{C}_{2}}$
The total number of ways to select 3 people from 20 people = ${}^{20}{{C}_{3}}$
$\Rightarrow $ Probability of selecting only 1 graduate = $\dfrac{{}^{5}{{C}_{1}}\times {}^{15}{{C}_{2}}}{{}^{20}{{C}_{3}}}$
(2) Here we assume that two people selected turn out to be graduated, which means one person selected is not graduated. So we have,
$\Rightarrow $ Number of ways to select 2 graduates from 5 graduates and 1 non graduate from 15 non graduates simultaneously = ${}^{5}{{C}_{2}}\times {}^{15}{{C}_{1}}$
The total number of ways to select 3 people from 20 people = ${}^{20}{{C}_{3}}$
$\Rightarrow $ Probability of selecting only 2 graduates = $\dfrac{{}^{5}{{C}_{2}}\times {}^{15}{{C}_{1}}}{{}^{20}{{C}_{3}}}$
(3) Here we assume that all the three people selected turns out to be graduated, so we have,
$\Rightarrow $ Number of ways to select 3 graduate from 5 graduates = ${}^{5}{{C}_{3}}$
The total number of ways to select 3 people from 20 people = ${}^{20}{{C}_{3}}$
$\Rightarrow $ Probability of selecting only 1 graduate = $\dfrac{{}^{5}{{C}_{3}}}{{}^{20}{{C}_{3}}}$
Now, since all the above three cases are possible, the total probability will be equal to the sum of probabilities of individual cases. Using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\Rightarrow $ Total probability = $\dfrac{{}^{5}{{C}_{1}}\times {}^{15}{{C}_{2}}}{{}^{20}{{C}_{3}}}+\dfrac{{}^{5}{{C}_{2}}\times {}^{15}{{C}_{1}}}{{}^{20}{{C}_{3}}}+\dfrac{{}^{5}{{C}_{3}}}{{}^{20}{{C}_{3}}}$
$\Rightarrow $ Total probability = $\dfrac{525}{20\times 19\times 3}+\dfrac{150}{20\times 19\times 3}+\dfrac{10}{20\times 19\times 3}$
$\therefore $ Total probability = $\dfrac{137}{228}$

So, the correct answer is “Option d”.

Note: Note that here you must not apply the formula for the permutations because we have to select people and not arrange them. If after selection we have to arrange the objects then we directly apply the formula of permutation given as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ without using the combinations formula. Remember the formula of probability as ‘number of favourable ways of selection to the total number of ways of selection’.