Question

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked randomly. What is the probability that it is neither blue nor green?
(a) \[\dfrac{2}{3}\]
(b) \[\dfrac{3}{4}\]
(c) \[\dfrac{7}{19}\]
(d) \[\dfrac{8}{21}\]

Answer 1

Hint: To solve this question, we need to know the concept of probability that is nothing but the ratio of the favorable number of outcomes to the total number of outcomes of an event. We can represent it mathematically as
\[\text{Probability = }\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\]

Complete step-by-step answer:

In this question, we have to find the probability of a ball that is picked up randomly and is neither blue nor green. To find this we need to know the concept of the probability that is nothing but the ratio of the favorable number of outcomes to the total number of outcomes of an event. Mathematically, we can represent it as,
 \[\text{Probability = }\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\]
To solve this, we will require the probability of choosing the blue ball which is
\[\text{P}\left( B \right)=\dfrac{\text{Number of blue balls }}{\text{Total number of balls}}\]
\[P\left( B \right)=\dfrac{7}{21}\]
\[P\left( B \right)=\dfrac{1}{3}\]
Also, we will require the probability of choosing the green ball which is
\[\text{P}\left( G \right)=\dfrac{\text{Number of green balls }}{\text{Total number of balls}}\]
\[P\left( G \right)=\dfrac{6}{21}\]
\[P\left( G \right)=\dfrac{2}{7}\]
Now, we have to find the probability of neither blue nor green balls. So, we can say that the probability is
P (Neither blue nor green) = 1 – P(B) – P(G)
Because the total probability of an event is always 1 and if we subtract the probabilities of events which we do not require from 1, we will get the required event probability. Now, we will put the values of P(B) and P(G). So, we will get,
P (Neither blue nor green) = \[1-\dfrac{1}{3}-\dfrac{2}{7}\]
Now, we will simplify it to get the answer.
P (Neither blue nor green) \[=\dfrac{21-7-6}{21}\]
P (Neither blue nor green) = \[\dfrac{8}{21}\]
Hence, the probability of the ball which we picked up is neither blue nor green is \[\dfrac{8}{21}\].
Therefore, option (d) is the right answer.

Note: We can also find the probability of choosing neither blue nor green ball by finding the probability of choosing the red ball because the box contains only 3 colors and out of the 3, we do not want 2 colors that are blue and green. So, definitely, we want the red color ball, that is \[P\left( R \right)=\dfrac{8}{21}\]= P (Neither blue nor green).