Question

A bag contains balls which are either red, white or blue.
There are twice as many red balls as white balls.
The probability of taking a blue ball from the bag is 1/ 10
What is the probability of taking a white ball from the bag?
A bag contains balls which are either red, white or blue.
There are twice as many red balls as white balls.
The probability of taking a blue ball from the bag is 1/ 10
What is the probability of taking a white ball from the bag?

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Let assume that total number of balls in a bag is y.

Further,

Assume that

➢ Number of white balls in a bag is 'x'.

So,

➢ Number of red balls in a bag is '2x'.

➢ Let number of blue balls be 'z'

It is given that,

➢ Probability of getting blue ball = 1/10

We know,

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

So,

$\rm :\longmapsto\:\dfrac{1}{10} = \dfrac{z}{y}$

$\rm :\implies\:y = 10z - - - (1)$

Also,

➢ Total number of balls = y

$\rm :\longmapsto\:x + 2x + z = y$

$\rm :\longmapsto\:3x + z = 10z$

$\rm :\longmapsto\:3x = 10z - z$

$\rm :\longmapsto\:3x = 9z$

$\rm :\implies\:x = 3z$

Now, we have

➢ Number of white balls, x = 3z

➢ Total number of balls, y = 10z

So, Probability of getting white ball is given by

$\rm :\longmapsto\:\sf \:P(getting \: white \: ball) =\dfrac{Number \: of \: white \: ball}{Total \: number \: of \: balls}$

$\rm :\longmapsto\:\sf \:P(getting \: white \: ball) =\dfrac{3z}{10z}$

$\rm :\longmapsto\:\bf \:P(getting \: white \: ball) =\dfrac{3}{10}$

Additional Information :-

Explore more :-

• The sample space associated with any random experiment is the collection of all possible outcomes.

• The probability of any outcome is a number between 0 and 1 including 0 and 1.

• The probability of sure event is 1.

• The probability of impossible event is 0.

• P(A) + P(A') = 1