Problem Statement
A bakery's croissants have $0.06$ defect probability.
(a) In a box of $20$, find $P(X = 2)$. [2 marks]
(b) In a box of $20$, find $P(X \ge 1)$. [2 marks]
(c) Find minimum $n$ such that $P(X \ge 1) > 0.99$. [2 marks]
Verified Solution & Marking Scheme
Part (a)
$P(X = 2) = \binom{20}{2}(0.06)^2(0.94)^{18} \approx 0.225$
Part (b)
$P(X \ge 1) = 1 - (0.94)^{20} \approx 0.710$
Part (c)
$1 - (0.94)^n > 0.99 \implies (0.94)^n < 0.01 \implies n > \frac{\ln 0.01}{\ln 0.94} \approx 74.43 \implies n_{\min} = 75$