Question

a box contains 7 white balls, 6 black balls, and 1 red balls.
a) how many red balls must be added to the box so that the probability of drawing a red ball is \\( \frac{4}{5} \\)?
b) how many black balls must be added to the original box so that the probability of drawing a white ball is \\( \frac{1}{5} \\)?
a) \\( \square \\) red balls would have to be added.
b) \\( \square \\) black balls would have to be added.

Explanation:

Part (a)

Step1: Define variables and total

Let \( x \) be red balls added. Original red: 1, total balls: \( 7 + 6 + 1 = 14 \). New red: \( 1 + x \), new total: \( 14 + x \). Probability: \( \frac{1 + x}{14 + x} = \frac{4}{5} \).

Step2: Solve equation

Cross - multiply: \( 5(1 + x)=4(14 + x) \)
\( 5 + 5x = 56 + 4x \)
\( 5x - 4x = 56 - 5 \)
\( x = 51 \).

Part (b)

Step1: Define variables and total

Let \( y \) be black balls added. Original white: 7, total balls: \( 14 + y \) (since only black added). Probability: \( \frac{7}{14 + y}=\frac{1}{5} \).

Step2: Solve equation

Cross - multiply: \( 7\times5 = 14 + y \)
\( 35 = 14 + y \)
\( y = 35 - 14 = 21 \).

Answer:

s:
a) \( \boldsymbol{51} \) red balls.
b) \( \boldsymbol{21} \) black balls.