Question

In a box, there are 8 red, 7 blues and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red or nor green?
A. \[\dfrac{4}{5}\]
B. \[\dfrac{2}{3}\]
C. \[\dfrac{1}{3}\]
D. None of the above

Answer 1

Hint: First find out the total number of balls, which is equal to the total number of possible events. Then according to the condition given that we have to find the probability that the ball drawn is neither red nor green is actually the probability of picking up a blue ball.

Complete step-by-step answer:

We have 8 red balls, 7 blue balls, 6 green balls
∴Total number of balls = S =8 + 7 + 6 = 21
Total number of possible outcomes= \[n\left( s \right)\] = 21
Let E = event that the ball drawn is neither red or green.
∴ n( E) = 7 (\[\because \] the ball drawn will be one of the 7 blue balls out of 21)
∴ number of favourable outcomes = 7
Using the formula probability= number of favourable outcomes/ total number of possible outcomes
\[ \Rightarrow P(E) = \dfrac{{n(E)}}{{n(S)}} = \dfrac{7}{{21}}\]
\[ \Rightarrow P(E) = \dfrac{1}{3}\].

Therefore, the correct option is ‘C’.

Note: In the above question, the event given is a mutually exclusive event. Mutually exclusive events are the ones that cannot occur at the same time. Here also we had to pick one at a time so one colour at a time.