Question

A box contains 3 red, 4 yellow and \((x+1)\) green balls. If two balls are taken out, then the probability that both the balls being green is \(\frac {5}{33}\). Find the value of x?

• A. 3
• B. 4
• C. 2
• D. 1

Answer 1

The Correct Option is B

To find the value of \(x\), we need to calculate the probability of drawing 2 green balls from the total number of balls and set it equal to \(\frac{5}{33}\).

Step 1: Determine the total number of balls. 
There are 3 red, 4 yellow, and \((x+1)\) green balls.
Total = \(3 + 4 + (x+1) = x + 8\).

Step 2: Calculate the probability of drawing 2 green balls.
The number of ways to choose 2 green balls from \((x+1)\) is \(\binom{x+1}{2} = \frac{(x+1)x}{2}\).
The total number of ways to choose any 2 balls from \(x+8\) is \(\binom{x+8}{2} = \frac{(x+8)(x+7)}{2}\).

Step 3: Set up the probability equation.
\[\frac{\binom{x+1}{2}}{\binom{x+8}{2}} = \frac{5}{33}\]
Substitute the combinations:
\[\frac{\frac{(x+1)x}{2}}{\frac{(x+8)(x+7)}{2}} = \frac{5}{33}\]
\[\frac{(x+1)x}{(x+8)(x+7)} = \frac{5}{33}\]

Step 4: Solve for \(x\).
Cross-multiply:
\[(x+1)x \cdot 33 = 5 \cdot (x+8)(x+7)\]
\[33x^2 + 33x = 5(x^2 + 15x + 56)\]
Expand and simplify:
\[33x^2 + 33x = 5x^2 + 75x + 280\]
Rearrange terms:
\[28x^2 - 42x - 280 = 0\]
Divide by 2:
\[14x^2 - 21x - 140 = 0\]

Step 5: Factor the quadratic equation.
To factor:
We need factors of \(14 \times -140 = -1960\) that add to \(-21\).
Solutions are \(x=4\) and \(x=-2.5\). Since \(x\) must be a positive integer, the solution is \(x=4\).

The value of \(x\) is 4.