Question

A box contains 4 white and 6 black balls. If 3 balls are drawn at random with replacement, what is the probability that at least one is white?

• A. \( \frac{27}{125} \)
• B. \( \frac{98}{125} \)
• C. \( \frac{64}{125} \)
• D. \( \frac{61}{125} \)

Hint:

To find the probability of at least one event occurring, calculate \( 1 - P(\text{none occur}) \), especially for independent events like draws with replacement.

Answer 1

The Correct Option is B

The probability of drawing a white ball is: \[ P(\text{white}) = \frac{4}{4 + 6} = \frac{4}{10} = \frac{2}{5} \] The probability of drawing a black ball is: \[ P(\text{black}) = \frac{6}{10} = \frac{3}{5} \] Since draws are with replacement, the probability of all three balls being black is: \[ P(\text{all black}) = \left( \frac{3}{5} \right)^3 = \frac{27}{125} \] The probability of at least one white ball is: \[ P(\text{at least one white}) = 1 - P(\text{all black}) = 1 - \frac{27}{125} = \frac{125 - 27}{125} = \frac{98}{125} \] The probability is: \[ \boxed{\frac{98}{125}} \]