Question

A bag contains 5 white, 7 red, and 4 black balls. Four balls are drawn one by one with replacement. The chance that at least two balls are black is

• A. \( \frac{67}{256} \)
• B. \( \frac{54}{256} \)
• C. \( \frac{243}{256} \)
• D. None of these

Hint:

To find the probability of at least a certain number of successes, calculate the complementary probability and subtract from 1.

Answer 1

The Correct Option is A

Step 1: Total Probability for Black Balls
The total number of balls is \( 5 + 7 + 4 = 16 \).
The probability of drawing a black ball on a single trial is: \[ P(\text{black}) = \frac{4}{16} = \frac{1}{4} \] The probability of not drawing a black ball is: \[ P(\text{not black}) = 1 - \frac{1}{4} = \frac{3}{4} \]
Step 2: Calculate the Probability of At Least Two Black Balls
The probability of drawing 0 or 1 black ball in 4 trials is easier to compute first, and we subtract it from 1 to get the probability of drawing at least 2 black balls. Probability of 0 black balls: \[ P(0 \text{ black balls}) = \left( \frac{3}{4} \right)^4 = \frac{81}{256} \] Probability of 1 black ball: \[ P(1 \text{ black ball}) = 4 \times \left( \frac{1}{4} \right) \times \left( \frac{3}{4} \right)^3 = 4 \times \frac{1}{4} \times \frac{27}{64} = \frac{108}{256} \]
Step 3: Final Calculation
Thus, the probability of drawing at least two black balls is: \[ P(\text{at least 2 black balls}) = 1 - \left( P(0 \text{ black balls}) + P(1 \text{ black ball}) \right) \] \[ P(\text{at least 2 black balls}) = 1 - \left( \frac{81}{256} + \frac{108}{256} \right) = 1 - \frac{189}{256} = \frac{67}{256} \]
Step 4: Conclusion
Thus, the correct answer is \( \frac{67}{256} \).