A bag contains 5 balls,8 white balls, 9 green balls,7 blue balls& 6 yellow balls.A ball is drawn at random .Find the probablity that the ball in neither green nor red ?

What color are the five balls mentioned first? You must have left out a word. Probably "red" . If a ball in neither green nor red, it must be white, yellow or blue. That probability is 21/35 = 3/5 = 60%

A bag contains 5 balls,8 white balls, 9 green balls,7 blue balls& 6 yellow balls.A ball is drawn at random .Find the probablity that the ball in neither green nor red ?

please let me know the formula also

You still have stated the problem incorrectly. You omitted the color of the first five balls. I have already answered what I the assume is the question

To find the probability that the ball is neither green nor red, we need to determine the total number of balls that are not green or red and divide it by the total number of balls in the bag.

Given that the bag contains:
- 5 balls
- 8 white balls
- 9 green balls
- 7 blue balls
- 6 yellow balls

To find the total number of balls that are not green or red, we need to exclude the green and red balls from the total number of balls.

Total number of balls = 5 (red + white + green + blue + yellow)
Total number of green and red balls = 9 green balls + 0 red balls = 9

Total number of balls that are not green or red = Total number of balls - Total number of green and red balls
= 5 - 9
= -4 (Since we cannot have a negative number of balls, we take it as 0)

Hence, the total number of balls that are neither green nor red is 0.

Now, we can calculate the probability.

Probability = Number of favorable outcomes / Total number of possible outcomes

In this case, the number of favorable outcomes (balls that are neither green nor red) is 0.

Therefore, the probability that the ball is neither green nor red is 0.