Question

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked at random. What is the probability that is neither red nor green?

Answer 1

Answer:

1/3

Step-by-step explanation:

no. of red balls=8

no. of blue balls=7

no. of green balls=6

total number of balls=21

number of balls that are neither red nor green=7

probability= number of favorable outcomes/total number of outcomes

probability= 7/21           [ on reducing }

probability=1/3

hope this helps, please mark brainliest

also.. follow me ♥

Answer 2

Given

• Number of red ball = 8
• Number of blue ball = 7
• Number of green ball = 6

To Find

• Probability that is neither red nor green.

Solution

★ Total number of balls

➞ 8 + 7 + 6

➞ 21

As we know that

$\sf P(E)=\dfrac{Number\:of\: favourable\: outcomes}{total\:number\:of\:possible\:outcomes}$

† Let E be the event of getting ball that is neither red or nor green.

$\therefore\sf P(E) = \dfrac{7}{21} = \dfrac{1}{3}$

Hence,

• Probability of getting ball that is neither red or nor green is 1/3

Additional Information

• P(E) = 0 {E is an impossible event}

• P(E) = 1 {E is a sure event}

$\rule{200}3$