Question

A bag contains 8 red, 6 white, and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is
(i) red or white (ii) not black (iii) neither white nor black $$$$

Answer 1

We find the total number of ways we can select 1 ball out of the total number of balls as sample size $n\left( S \right)$ using combination. We find the number of ways we can get ball that is red or white as $n\left( {{A}_{1}} \right)$, not black as $n\left( {{A}_{2}} \right)$ , neither white nor black as $n\left( {{A}_{3}} \right)$. We find the required probabilities$P\left( {{A}_{1}} \right)=\dfrac{n\left( {{A}_{1}} \right)}{n\left( S \right)},P\left( {{A}_{2}} \right)=\dfrac{n\left( {{A}_{2}} \right)}{n\left( S \right)},P\left( {{A}_{3}} \right)=\dfrac{n\left( {{A}_{3}} \right)}{n\left( S \right)}$. $$$$

Complete step-by-step solution:
We know from the definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring (or number of favorable 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 a bag contains 8 red, 6 white, and 4 black balls. A ball is drawn at random from the bag. The total number of balls is $8+6+4=18$. We can select 1 ball out of 18 in ${}^{18}{{C}_{1}}$ ways which is the number of all possible outcomes. So we have,
$n\left( S \right)={}^{18}{{C}_{1}}=\dfrac{18}{1}=18$
(i) We are asked to find the probability that the drawn ball is red or white. Let us denote the event that the drawn ball is red or white as${{A}_{1}}$. We can select 1 red ball out of 8 in ${}^{8}{{C}_{1}}$ way and select 1 white ball out of 6 ball in ${}^{6}{{C}_{1}}$way. We use the rule of sum and find the number of ways ${{A}_{1}}$ can occur as
$n\left( {{A}_{1}} \right)={}^{8}{{C}_{1}}+{}^{6}{{C}_{1}}=8+6=14$
So the required probability is
$P\left( {{A}_{1}} \right)=\dfrac{n\left( {{A}_{1}} \right)}{n\left( S \right)}=\dfrac{14}{18}=\dfrac{7}{9}$
(ii) We are asked to find the probability that the drawn ball is not black. Let us denote the event that the drawn ball is not black as${{A}_{2}}$. If the ball is not black it is either red or white for which we have obtained the probability in part(i). So the required probability is
$P\left( {{A}_{2}} \right)=P\left( {{A}_{1}} \right)=\dfrac{7}{9}$
(iii) We are asked to find the probability that the drawn ball is not neither white nor black. Let us denote the event that the drawn ball is neither white nor black as${{A}_{3}}$. If the ball is neither white nor black then the ball is red. We select 1 red ball out of 8 red balls in ${}^{8}{{C}_{1}}$ which is the number of ways ${{A}_{3}}$ can occur. So we have
$n\left( {{A}_{3}} \right)={}^{8}{{C}_{1}}=8$
So the required probability is
$P\left( {{A}_{3}} \right)=\dfrac{n\left( {{A}_{3}} \right)}{n\left( S \right)}=\dfrac{8}{18}=\dfrac{4}{9}$

Note: We note that the question presumes that all balls are equally likely to be picked. We can also find $P\left( {{A}_{2}} \right),P\left( {{A}_{3}} \right)$ with method of negation where first find the probability of getting a black ball$P\left( B \right)$, a white ball $P\left( W \right)$ and have $P\left( {{A}_{2}} \right)=1-P\left( B \right)$ . We also have $P\left( {{A}_{3}} \right)=1-\left( P\left( B \right)+P\left( W \right) \right)$ since getting a black or white ball are mutually exclusive events.