A bag contains 14 balls of which \(x\) are white. If 6 more white balls are added to the bag, the probability of drawing a white ball is \( \frac{1}{2} \). Then the value of \( x = \)
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1. Initial state: Total = 14, White = \(x\).
2. After adding 6 white balls:
New White = \(x+6\).
New Total = \(14+6 = 20\).
3. Probability of drawing a white ball now = (New White) / (New Total).
Given this probability is \(1/2\).
4. Equation: \( \frac{x+6}{20} = \frac{1}{2} \).
5. Solve for \(x\):
\(2(x+6) = 20 \times 1\)
\(2x + 12 = 20\)
\(2x = 20 - 12 = 8\)
\(x = 4\).
Concept: Probability = (Number of favorable outcomes) / (Total number of possible outcomes).
Step 1: Describe the initial state of the bag
Initially:
Total number of balls = 14.
Number of white balls = \(x\).
Number of non-white balls = \(14 - x\).
Step 2: Describe the state of the bag after adding more white balls
6 more white balls are added to the bag.
After adding:
New number of white balls = \(x + 6\).
New total number of balls = Initial total + Added balls = \(14 + 6 = 20\).
Step 3: Set up the probability equation for drawing a white ball after adding more
The probability of drawing a white ball from the modified bag is given as \( \frac{1}{2} \).
Using the probability formula:
P(drawing a white ball) = \(\frac{\text{New number of white balls}}{\text{New total number of balls}}\)
\[ \frac{x+6}{20} = \frac{1}{2} \]
Step 4: Solve the equation for \(x\)
\[ \frac{x+6}{20} = \frac{1}{2} \]
To solve for \(x\), we can cross-multiply or multiply both sides by 20.
Multiplying both sides by 20:
\[ x+6 = \frac{1}{2} \times 20 \]
\[ x+6 = 10 \]
Subtract 6 from both sides:
\[ x = 10 - 6 \]
\[ x = 4 \]
Step 5: Check the answer (optional)
If \(x=4\):
Initially: 4 white balls, total 14 balls.
After adding 6 white balls:
Number of white balls = \(4+6 = 10\).
Total number of balls = \(14+6 = 20\).
Probability of drawing a white ball = \(\frac{10}{20} = \frac{1}{2}\).
This matches the given information.
Therefore, the value of \(x\) is 4.