Question

Consider a bag of 10 balls containing 3 white, 4 black, 2 red and one green balls . A ball is drawn at random . What is the probability of drawing (a) a white ball (b) green ball (c) black ball (d) red ball ?
Step by Step explanation : ​

Answer 1

Answer

5

White

6

Red

9

Green

Total number of balls =5+6+9

=20

(i) Probability =

Total cases

favourable cases

P(Green) =

20

9

(ii) P(White or red) =

20

5+6

=

20

11

(iii) P(neither green nor white) = P(Red)

=

20

6

=

10

3

Answer 2

$\mathbb{\huge{\green{Your\: Answer:}}}$

a. Probability of getting white ball =

$= \frac{3}{10}$

b. Probability of getting green ball =

$= \frac{1}{10}$

c. Probability of getting black ball =

$= \frac{4}{10} = \frac{2}{5}$

d. Probability of getting red ball =

$\frac{2}{10} = \frac{1}{5}$

Here is a TIP:

• Probability of anything:

$\frac{ Number \: of \: the \: thing}{Total \: number \: of \: all \: things}$

Explanation:

• Total number of balls = 10
• Number of White ball = 3
• Number of green ball = 1
• Number of black ball = 4
• Number of Red ball = 2

Step-by-step Solution:

a. Probability of getting white ball =

$= \frac{3}{10}$

b. Probability of getting green ball =

$= \frac{1}{10}$

c. Probability of getting black ball =

$= \frac{4}{10} = \frac{2}{5}$

d. Probability of getting red ball =

$\frac{2}{10} = \frac{1}{5}$