Question

A bag contains 9 red balls and some blue balls. If the probability of drawing a blue ball is three times of
that of a red ball, then find the total number of balls in the bag.

Answer 1

Answer:

Let number of blue balls in the bag =x

Total no of balls in bag =5+x [No. of redball =5]

Probability of drawing a blue ball =

Total no of ball

No. of blue ball

P(B)=

5+x

x

Probability of drawing a Red ball =

Total no of ball

No. of red ball

P(R)=

5+x

5

Given,

P(B)=2P(R)

5+x

x

=2(

5+x

5

)

x=10

Hence, no. of blue balls in the bag =10.

Answer 2

Given :

• A bag contains 9 red balls and some blue balls. The probability of drawing a blue ball is three times of that of a red ball.

To Find :

• The total numbers of balls in the bag = ?

Answer :

• Let the number of blue ball be 'x'.

• Number of red ball = 9

• Therefore,total number balls = blue balls + red balls = x + 9

• $\sf P\: (getting\: red\: ball) =\dfrac{No. \:of\: red \:balls}{Total \:no. \:of\: balls\: in\: bag} = \dfrac{9}{x + 9}$

• $\sf P \:(getting \:blue\: ball) = \dfrac{No. \:of\: blue\:balls}{Total \:no. \:of\: balls\: in\: bag} = \dfrac{x}{x + 9}$

★ According to Question now,

➝ $\sf 3 \left( \dfrac{9}{x + 9} \right) = \dfrac{x}{x + 9}$

➝ $\sf 27 \left( x + 9 \right) = x^2 + 9x$

➝ $\sf 27x + 243 = x^2 + 9x$

➝ $\sf x^2 + 9x - 27x - 243 = 0$

➝ $\sf x^2 - 18x - 243 = 0$

➝ $\sf x^2 - 27x + 9x - 243 = 0$

➝ $\sf x(x - 27) + 9(x - 27) = 0$

➝ $\sf (x + 9) (x - 27) = 0$

➝ $\sf x = -9\: or\: x = 27$

As we know that, number of balls cannot be negative.

⛬ x = 27

Hence, Number of blue balls is 27.

Therefore,

• Total number of balls in bag = x + 9 = 27 + 9 = 36