Question

In a box, there are eight red, seven blue and six green balls. One ball is picked randomly. What is the probability that it is neither red nor green?

Answer 1

Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Define E as the event: Ball drawn is green and F the event ball drawn is red. Find the probability $P\left( E\bigcup F \right)$ using the fact that if E and F are mutually exclusive events, then $P\left( E\bigcup F \right)=P\left( E \right)+P\left( F \right)$. Use De-morgan's law and the fact that $P\left( A' \right)=1-P\left( A \right)$ to prove that $P\left( E'\bigcap F' \right)=1-P\left( EUF \right)$ and hence find the probability $P\left( E'\bigcap F' \right)$, which is the required probability that the ball chosen is neither red nor green.

Let E be the event: The ball chosen is red
Let F be the event: The ball chosen is green

Complete step-by-step answer:
Since there are eight red balls, the number of cases favourable to E is 8.
Hence, we have n (E) = 8
Since there are six green balls, the number of cases favourable to F is 6.
Hence, we have n(F) = 6
The total number of ways in which we can choose a ball = 21
Hence, we have n (S) = 21
Hence, P (E) = $\dfrac{8}{21}$ and P(F) $=\dfrac{6}{21}$
Since E and F are mutually exclusive events(because a ball cannot be both red and green), we have
$P\left( E\bigcup F \right)=P\left( E \right)+P\left( F \right)=\dfrac{8}{21}+\dfrac{6}{21}=\dfrac{14}{21}$
Now, we know from de-morgan’s laws \[E'\bigcap F'=\left( E\bigcup F \right)'\]
Hence, we have $P\left( E'\bigcap F' \right)=P\left( \left( E\bigcup F \right)' \right)$
We know that $P\left( A' \right)=1-P\left( A \right)$
Hence, we have
$P\left( E'\bigcap F' \right)=1-P\left( E\bigcup F \right)=1-\dfrac{14}{21}=\dfrac{7}{21}=\dfrac{1}{3}$
Hence the probability that the chosen ball is neither red nor green is $\dfrac{1}{3}$

Note: [1] It is important to note that drawing uniformly at random is important for the application of the above problem. If the draw is not random, then there is a bias factor in drawing, and the above formula is not applicable. In those cases, we use the conditional probability of an event.
[2] Note that the event that the drawn ball is neither green nor red is equivalent to the event ball drawn is blue. Hence, we have the probability that the ball drawn is neither red nor green is $\dfrac{7}{21}=\dfrac{1}{3}$, which is the same as obtained above.
[3] The sum of probabilities of an event E and its complement E’ = 1
i.e. $P(E)+P(E')=1$
Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(E’) instead of P(E).