A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting neither a green ball nor a red ball.

# A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting neither a green ball nor a red ball.

1. A
$\frac{17}{20}$
2. B
$\frac{7}{20}$
3. C
$\frac{1}{20}$
4. D
$\frac{3}{20}$

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

Given that there are 5 red, 7 white and 8 green balls in a bag.
We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total possible outcomes n}\left(S\right)}$
Total number of balls are:
$5+7+8=20$
$⇒n\left(S\right)=20$
Let E be the event of getting neither a green nor a red ball.
The probability of getting neither a green nor a red ball is equal to the probability of getting a white ball.
Total number of white balls is 7.
$⇒n\left(E\right)=7$
So the required probability is,
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total number of outcomes n}\left(S\right)}$
$⇒P\left(E\right)=\frac{7}{20}$
Thus, the probability of getting neither green nor a red ball is $\frac{7}{20}$.
Hence, option 2 is correct.

## Related content

 Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)