A bucket contains 10 brown balls, 8 green balls, and 12 red balls and one ball is picked randomly without looking. The probability that the picked ball will be brown is _______.

$\frac{4}{15}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1}{3}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.61

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$\frac{1}{3}$

Total number of balls = 10 + 8 + 12 = 30 = n(S)

Total number of brown balls = 10 = n(B)

Let the probability of the picked brown ball be P(B).

Therefore,

P(B) = $\frac{n (B)}{n (S)} = \frac{10}{30} = \frac{1}{3} = 0.33$ .

Suggest Corrections

Similar questions

A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one without looking. What is the probability that the ball will be brown?

Q. A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is

Q. 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?

Q. A box contains

7

white,

5

red and

3

black balls. Three balls are picked randomly one at a time, and are being replaced. What is the probability of picking a white then red then a black ball?

Q. An urn contains 9 red balls and x green balls. If the probability of picking a red ball is thrice that of picking a green ball, then x = ________.

Join BYJU'S Learning Program

A bucket contains 10 brown balls, 8 green balls, and 12 red balls and one ball is picked randomly without looking. The probability that the picked ball will be brown is _______.

The correct option is B 13 Total number of balls = 10 + 8 + 12 = 30 = n(S) Total number of brown balls = 10 = n(B) Let the probability of the picked brown ball be P(B). Therefore, P(B) = n(B)n(S)=1030=13=0.33.

The correct option is B
$\frac{1}{3}$

Total number of balls = 10 + 8 + 12 = 30 = n(S)

Total number of brown balls = 10 = n(B)

Let the probability of the picked brown ball be P(B).

Therefore,

P(B) = $\frac{n (B)}{n (S)} = \frac{10}{30} = \frac{1}{3} = 0.33$ .