An urn contains 4 red , 5 green , 6 blue and some yellow balls . If two balls are drawn at random ,the probability of getting at least one yellow ball is `(17)/(38)` .find the yellow balls in the urn.

Options

A. 4
B. 5
C. 6
D. 10

Answer

To solve the problem step by step, let's break down the information given and the calculations needed to find the number of yellow balls in the urn.

### Step 1: Understand the Problem
We have an urn containing:
- 4 red balls
- 5 green balls
- 6 blue balls
- y yellow balls (unknown quantity)

The total number of balls in the urn is \( 4 + 5 + 6 + y = 15 + y \).

### Step 2: Define the Probability of Drawing Balls
We need to find the probability of drawing at least one yellow ball when two balls are drawn at random. The probability given is \( \frac{17}{38} \).

### Step 3: Calculate the Total Outcomes
The total number of ways to choose 2 balls from \( 15 + y \) balls is given by the combination formula:
\[ \text{Total Outcomes} = \binom{15 + y}{2} = \frac{(15 + y)(14 + y)}{2} \]

### Step 4: Calculate the Favorable Outcomes
To find the probability of getting at least one yellow ball, we can use the complement rule. First, we will calculate the probability of getting no yellow balls.

The number of ways to choose 2 balls from the non-yellow balls (which are 15 in total) is:
\[ \text{No Yellow Outcomes} = \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

### Step 5: Probability of No Yellow Balls
The probability of drawing no yellow balls is:
\[ P(\text{No Yellow}) = \frac{\text{No Yellow Outcomes}}{\text{Total Outcomes}} = \frac{105}{\frac{(15 + y)(14 + y)}{2}} = \frac{210}{(15 + y)(14 + y)} \]

### Step 6: Probability of At Least One Yellow Ball
The probability of getting at least one yellow ball is:
\[ P(\text{At least one Yellow}) = 1 - P(\text{No Yellow}) = 1 - \frac{210}{(15 + y)(14 + y)} \]

Setting this equal to the given probability:
\[ 1 - \frac{210}{(15 + y)(14 + y)} = \frac{17}{38} \]

### Step 7: Solve the Equation
Rearranging the equation gives:
\[ \frac{210}{(15 + y)(14 + y)} = 1 - \frac{17}{38} = \frac{21}{38} \]

Cross-multiplying:
\[ 210 \cdot 38 = 21 \cdot (15 + y)(14 + y) \]
\[ 7980 = 21(210 + 29y + y^2) \]
\[ 7980 = 4410 + 609y + 21y^2 \]
\[ 21y^2 + 609y - 3570 = 0 \]

### Step 8: Simplify the Quadratic Equation
Dividing the entire equation by 3:
\[ 7y^2 + 203y - 1190 = 0 \]

### Step 9: Use the Quadratic Formula
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 7, b = 203, c = -1190 \):
\[ b^2 - 4ac = 203^2 - 4 \cdot 7 \cdot (-1190) \]
\[ = 41209 + 33340 = 74549 \]
\[ y = \frac{-203 \pm \sqrt{74549}}{14} \]
Calculating \( \sqrt{74549} \approx 273 \):
\[ y = \frac{-203 \pm 273}{14} \]

Calculating the two possible values:
1. \( y = \frac{70}{14} = 5 \)
2. \( y = \frac{-476}{14} \) (not valid since y cannot be negative)

Thus, the number of yellow balls \( y = 5 \).

### Final Answer
The number of yellow balls in the urn is **5**.