Question

In a bag there are 21 balls some red and some green. The probability of drawing a red ball is known to be 1/3.
a) How many red balls are there in the bag?
b) If one red ball is taken out, what would be the probability of drawing a red ball from the remaining balls?​

Answer 1

Answer:

a)There are 7 red balls in the bag because 1/3 of 21 is 7

b)Probability of drawing red ball from remaining ball =

6÷21=3 whole and 1/2

Answer 2

Given

$\bold{Total \ no: \ of \ balls} = 21$

Let x and y be the total no: of red and green balls respectively.

a)

$\bold{Probability \ of \ getting \ a \ red \ ball} = \frac{1}{3}$

$\frac{1}{3} = \frac{x}{21} \\=> x = 21/3\\=> x = 7$

$\boxed{\boxed{\bold{Total \ no : \ of \ red \ balls} = 7}}}}$

b)

$\bold{ One \ ball \ is \ taken \ out} \\=> \bold{ Total \ no: \ of \ red \ balls} = 6$

$\bold{ Total \ no : \ of \ balls } = 20$

$=> \bold{Probability \ of \ drawing \ a \ red \ ball =\boxed{\boxed{\bold{ \frac{6}{20}}}}}$