<p>A bag contains 3 white and 5 black balls. One ball is drawn at random and then replaced. The process is repeated 4 times. The probability that at least one white ball is drawn is</p>
Step-by-Step Solution
Key Concept: Use complement: P(at least one white) = 1 - P(no white). Note options A and B are equal; B is the more natural form.
<p>$P(\text{white}) = \dfrac{3}{8}$, $P(\text{black}) = \dfrac{5}{8}$.</p><p>$P(\text{no white in 4 draws}) = \left(\dfrac{5}{8}\right)^4 = \dfrac{625}{4096}$... </p><p>Wait -- 4 draws with replacement from 8 balls: $P(\text{no white}) = \left(\dfrac{5}{8}\right)^4$.</p><p>$P(\text{at least 1 white}) = 1 - \left(\dfrac{5}{8}\right)^4 = \dfrac{4096-625}{4096} = \dfrac{3471}{4096}$.</p><p>Answer: $1 - \left(\dfrac{5}{8}\right)^4$.</p>
Correct Answer: B