Question

In a box, there are 4 red, 3 blue and 5 green balls. one ball is picked up randomly. what is the probability that it is neither red nor green?

Answer 1

neither red nor green is. blue
=3/12
=1/4

Answer 2

Answer:

The probability that the ball is neither red nor green is 1/4.

Step-by-step explanation:

Given 4 red, 3 blue and 5 green balls.

Let’s assume:

Number of red balls be n(R) which is n(R) =4;

Number of red blue be n(B) which is n(B)=3;

Number of red green be n(G) which is n(G)=5;

Let n(s) be no. of red, blue and green balls respectively,  

So n(s) =n(R)+n(B)+n(G)

Substitute the values to the n(s);

i.e n(s)=4+3+5;

n(s) =12;

From then question, we need to find the P(B) i.e probability of finding blue balls.

So, here P(neither red nor green) implies nothing but P(blue balls)

P(B)=n(B)n(S);

P(B)=3/12

P(B)=1/4;