Question

a bag contains 5 red 4 blue and 3 Green balls A ball is taken out of the bag at random find the probability that selected ball is of red colour and
not of green colour

Answer 1

TOTAL NO. OF BALL IN BAG= 5+4+3=12

1)P(RED COLOUR BALL) = 5/12

2)P(NOT GREEN COLOUR) = 5+4=9
=9/12=3/4.

Answer 2

Answer:

The probability of red color ball is ball is $\frac{5}{12}$

The probability of balls except green color is $\frac{3}{4}$

Step-by-step explanation:

Given: A bag contains 5 red 4 blue and 3 Green balls

To find: The probability of red color ball and probability of not of green color ball

Solution:

No of red color balls = 5

No. of blue color balls = 4

No of green color balls = 3

Total number of balls = 12

Therefore

n(S) = 12

Let A be the red balls taken out of the bag randomly.

n(A) = 5

Therefore,

⇒ $p(A) = \frac{n(A)}{n(S)}$

⇒ $p(A) = \frac{5}{12}$

The probability of red color ball is $\frac{5}{12}$

Let B be the favorable outcomes not of green color

Therefore total number of balls = 12

Total number of green balls = 3

Except green balls = 12 - 3

                              = 9

n(B) = 9

Therefore,

⇒ $p(B) = \frac{n(B)}{n(S)}$

⇒ $p(B) = \frac{9}{12}$

⇒ $p(B) = \frac{3}{4}$

The probability of balls except green color is $\frac{3}{4}$

Final answer:

The probability of red color ball is $\frac{5}{12}$ and the probability of ball except green color is $\frac{3}{4}$

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