In a bag ,ther are some red , black and yellow balls . Sum of black and yellow balls is 9 . Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even.

Options

A. 3
B. 5
C. 6
D. 4

Answer

To solve the problem step by step, we will break down the information provided and use the concepts of probability and combinations.

### Step 1: Define Variables
Let:
- \( r \) = number of red balls
- \( b \) = number of black balls
- \( y \) = number of yellow balls

From the problem, we know:
1. \( b + y = 9 \) (the sum of black and yellow balls is 9)
2. The probability of selecting two red balls is \( \frac{1}{7} \).
3. This probability is 250% of the probability of selecting two black balls.

### Step 2: Express the Probability of Selecting Two Red Balls
The probability of selecting 2 red balls from a total of \( r + b + y \) balls is given by:
\[ P(\text{2 red}) = \frac{\binom{r}{2}}{\binom{r + b + y}{2}} = \frac{\frac{r(r-1)}{2}}{\frac{(r + b + y)(r + b + y - 1)}{2}} \]
This simplifies to:
\[ P(\text{2 red}) = \frac{r(r-1)}{(r + b + y)(r + b + y - 1)} \]
Setting this equal to \( \frac{1}{7} \):
\[ \frac{r(r-1)}{(r + b + y)(r + b + y - 1)} = \frac{1}{7} \]

### Step 3: Express the Probability of Selecting Two Black Balls
The probability of selecting 2 black balls is:
\[ P(\text{2 black}) = \frac{\binom{b}{2}}{\binom{r + b + y}{2}} = \frac{\frac{b(b-1)}{2}}{\frac{(r + b + y)(r + b + y - 1)}{2}} \]
This simplifies to:
\[ P(\text{2 black}) = \frac{b(b-1)}{(r + b + y)(r + b + y - 1)} \]

### Step 4: Relate the Probabilities
According to the problem, the probability of selecting two red balls is 250% of the probability of selecting two black balls:
\[ \frac{1}{7} = 2.5 \times P(\text{2 black}) \]
This means:
\[ P(\text{2 black}) = \frac{1}{2.5 \times 7} = \frac{1}{17.5} \]

### Step 5: Set Up the Equations
From the above probabilities, we can set up the following equations:
1. \( r(r-1) = \frac{1}{7} \times (r + b + y)(r + b + y - 1) \)
2. \( b(b-1) = \frac{1}{17.5} \times (r + b + y)(r + b + y - 1) \)

### Step 6: Solve for \( r \)
Using the first equation:
\[ 7r(r-1) = (r + b + y)(r + b + y - 1) \]
Substituting \( b + y = 9 \) into \( r + b + y \):
\[ 7r(r-1) = (r + 9)(r + 8) \]
Expanding both sides and simplifying will give us a quadratic equation in terms of \( r \).

### Step 7: Solve for \( b \) and \( y \)
After finding \( r \), substitute back into the equations to find \( b \) and \( y \). Remember that \( b \) is even.

### Step 8: Conclusion
After solving the equations, we find the values of \( r \), \( b \), and \( y \).

### Final Result
After calculations, we find that the number of yellow balls \( y \) is **5**.

---