Question

From a group of 3 mathematicians, 4 statisticians, and 5 economists, a committee of 4 is to be selected by lottery. So, find the probability that the committee consists of 4 economists.
A. $\dfrac{{}^{7}{{C}_{4}}}{{}^{12}{{C}_{4}}}$
B. $\dfrac{{}^{4}{{C}_{4}}}{{}^{12}{{C}_{4}}}$
C. $\dfrac{{}^{5}{{C}_{4}}}{{}^{12}{{C}_{4}}}$
D. $\dfrac{{}^{13}{{C}_{2}}}{{}^{12}{{C}_{4}}}$

Answer 1

Hint: We will first start by defining the combination formula that is how we select objects. First, we will select a total of 4 people out of all the 12 persons and then 4 economists out of 5 economists, and then we will apply the probability formula that is: $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, to find the probability that the selected 4 members are 4 economists.

Complete step-by-step solution
First, we will see what is meant by the combination formula. So, the combination formula is used to find the number of ways of selecting items from a collection, such that the order of selection does not matter. In simple words, combination involves the selection of objects or things out of a larger group where order doesn’t matter.
So, for selecting r objects out of a total of n objects, the combination formula is as follows:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ ,
Where n and r are non-negative integers and $!$ is the factorial operator.
So, we are given that we have 3 mathematicians, 4 statisticians, and 5 economists that means a total of $3+4+5=12$ persons and now, we have to select a committee of 4 out of 12 so we will apply the combination formula that is: ${}^{12}{{C}_{4}}$
Now, we have to select 4 economists out of the 5 economists so we will again apply the combination formula: ${}^{5}{{C}_{4}}$
Now, we will apply the formula for the probability that is: $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Now, our favorable outcomes are ${}^{5}{{C}_{4}}$ that is selecting four economists out of 5 and the total number of outcomes is ${}^{12}{{C}_{4}}$ that is selecting 4 people out of 12:
Therefore, the probability that the committee consists of 4 economists is $\dfrac{{}^{5}{{C}_{4}}}{^{12}{{C}_{4}}}$.
Hence, the correct answer is C.

Note: If we were asked to expand the answer further or the options were given in the numerical form, we would apply the combination formula that is: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, so: $\dfrac{{}^{5}{{C}_{4}}}{^{12}{{C}_{4}}}=\dfrac{\dfrac{5!}{4!\left( 5-4 \right)!}}{\dfrac{12!}{4!\left( 12-4 \right)!}}=\dfrac{5!}{4!\left( 5-4 \right)!}\times \dfrac{4!\left( 12-4 \right)!}{12!}=\dfrac{5!8!}{1!12!}=\dfrac{5!8!}{8!\left( 9\times 10\times 11\times 12 \right)}=\dfrac{120}{120\times 9\times 11}=\dfrac{1}{99}$
A common mistake made is by directly finding out the probability of having 4 economists out of 12 people that is $\dfrac{4}{12}=\dfrac{1}{3}$, here no selection is made hence, the answer is wrong.