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?

Answer 1

No. of favourable outcome=8+7+6=21
No. of possible outcome=7
P(getting a ball which is neither red nor green)=7/21=1/3(Ans)
Hope this answer helps you.

Answer 2

Given :

• No. of Red balls = 8
• No. of Blue balls = 7
• No. of Green balls = 6

To Find :

• Probability that is neither red nor blue.

Solution :

$\longmapsto\tt{Total\:balls=8+7+6}$

$\longmapsto\tt\bold{21\:balls}$

Total balls - red balls - blue balls

$\longmapsto\tt{21-8-6}$

$\longmapsto\tt\bold{7\:balls}$

Now :

$\tt\boxed{Probability=\dfrac{No.\:of\:fav.\:Outcomes}{Total\:no.\:of\:Outcomes}}$

$\longmapsto\tt{\cancel\dfrac{7}{21}}$

$\longmapsto\tt\bold{Probability=\dfrac{1}{3}}$

So , The probabilty of getting neither red nor green ball is 1/3..