Question

There are 5 positive numbers and 6 negative numbers. Three numbers are chosen at random and multiplied. The probability that the product being a negative number is $$$$
A.\dfrac{11}{34}$$$$$ B. \dfrac{17}{33} C.$\dfrac{16}{55}
D.\dfrac{15}{34}$$$$$ E.\dfrac{16}{33}$$$$$

Answer 1

We find the number of ways we can select 3 numbers out total given numbers as the total number of outcomes$n\left( S \right)$. We find the number of favourable outcomes as number of ways we choose 3 numbers whose product will be negative as $n\left( A \right)={{n}_{1}}\left( A \right)+{{n}_{2}}\left( A \right)$ where ${{n}_{1}}\left( A \right)$ is the number of ways we can choose 2 positive and negative numbers and ${{n}_{2}}\left( A \right)$ is the number of ways we can choose 3 negative numbers. The required probability is $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.$$$$

Complete step by step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring (or number of favourable outcomes) and $n\left( S \right)$ is the size of the sample space (number of all possible outcomes) then the probability of the event $A$ occurring is given by
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
We are given in the question that there are 5 positive numbers and 6 negative numbers. We have to choose 3 numbers and then multiply. So the total number of numbers we can choose is $5+6=11$. We can choose 3 numbers randomly out of 11 numbers in ${}^{11}{{C}_{3}}$ways. So the number of all possible outcomes is
$n\left( S \right)={}^{11}{{C}_{3}}=\dfrac{11!}{3!8!}=\dfrac{11\times 10\times 9}{1\times 2\times 3}=165$
We want the product of the chosen three numbers to be negative. Let us denote the event of getting a negative product as $A$. We have two cases here.
Case-1: We select 2 positive numbers whose product will be positive and 1 negative number with whom we multiply the product of two positive numbers to get the final product negative. We can choose 2 positive numbers out of given 5 positive numbers in ${}^{5}{{C}_{2}}$ way and 1 negative numbers out of 6 negative numbers in $^{6}{{C}_{1}}$ way. So by rule of product the number of ways we can choose following case-1 is
${{n}_{1}}\left( A \right)={}^{5}{{C}_{2}}\times {}^{6}{{C}_{1}}=10\times 6=60$
Case-2: We select three negative numbers where two negative numbers when multiplied give positive and with which we multiply the third negative number to get negative. We can select 3 negative numbers out of 6 negative numbers in ${}^{6}{{C}_{3}}$ ways. So the number of ways we can choose following case-2 is
${{n}_{2}}\left( A \right)={}^{6}{{C}_{3}}=\dfrac{6!3!}{3!}=\dfrac{6\times 5\times 4}{1\times 2\times 3}=20$
We can follow either case-1 or case-2 to get the negative product. So by rule of sum the number of favourable outcomes is number of ways we choose 3 numbers whose product will be negative is
$n\left( A \right)={{n}_{1}}\left( A \right)+{{n}_{2}}(A)=60+20=80$
So the required probability is
$P\left( A \right)=\dfrac{80}{165}=\dfrac{16}{33}$

So, the correct answer is “Option E”.

Note: We note that since the numbers are distinct mathematical objects we have used combination formula to select $r$ objects from $n$distinct objects as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We can alternatively solve by negation where we find the total number for ways we can select 3 numbers such that their product is positive as ${}^{5}{{C}_{3}}+{}^{6}{{C}_{2}}\cdot {}^{5}{{C}_{1}}$ and then subtract from ${}^{11}{{C}_{3}}$to get the number of favourable outcomes.