Question

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
$
  (a){\text{ }}\dfrac{1}{3} \\
  (b){\text{ }}\dfrac{3}{4} \\
  (c){\text{ }}\dfrac{7}{{19}} \\
  (d){\text{ }}\dfrac{8}{{21}} \\
  (e){\text{ }}\dfrac{9}{{21}} \\
$

Answer 1

Hint – In this question if the ball drawn is neither green nor red this means we are talking about the only remaining color in the box which is blue. Use the basic probability formula of favorable outcome divided by total outcome to get the answer.

Complete step-by-step answer:

Given data
Number of red ball in a box = 8
Number of blue ball in a box = 7
Number of green ball in a box = 6

So the total balls in a box = (8 + 7 + 6) = 21

So the total number of outcomes = 21.

Now one ball is picked up randomly and we have to find the probability that the ball is neither red nor green.

Now if the ball is neither red nor green than the remaining color left in the box is blue.
So the favorable number of outcomes are 7.

Now we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes.

So the probability (P) that the ball is neither red nor green is
$ \Rightarrow P = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{7}{{21}} = \dfrac{1}{3}$.
So this is the required probability.
Hence option (A) is correct.

Note – In mathematical terms if A is the event of drawing a red ball and B is the event of drawing a green ball than probability that neither a red ball nor a green ball is drawn means we are talking about \[P\left( {\overline A \cap \overline B } \right)\]. We could have also used this concept to solve these problems but it was time taking, thus it is advised to use general mathematics approach to such probability questions.