A-Level Discrete random variables: worked solution

Question

Now p = 5. Boxes are packed 12 per carton; a carton is accepted if at most 1 box is seasonal, else rejected. Given a carton is rejected, find the probability that at most 30% of its boxes are seasonal.

Method: turn "at most 30% of its boxes" into a whole number of boxes first, then use the conditional probability formula. The condition "rejected" restricts the sample space, so the answer is not just P(X <= 3). Step 1: Set up the distribution. p = 5 means 5%, so each box is seasonal with probability 0.05, and there are 12 boxes per carton. Let X be the number of seasonal boxes in a carton: X ~ B(12, 0.05) Step 2: Translate the two events. Rejected means MORE than 1 seasonal box, i.e. X >= 2. "At most 30% of its boxes" means at most 0.30 x 12 = 3.6 boxes; X counts whole boxes, so this means X <= 3. Step 3: Use P(A | B) = P(A and B) / P(B), with A: X <= 3 and B: X >= 2. A and B together mean 2 <= X <= 3. Required probability = P(2 <= X <= 3) / P(X >= 2) Step 4: Work out each piece from the binomial formula P(X = r) = C(12, r)(0.05)^r (0.95)^(12-r). P(X = 0) = (0.95)¹² = 0.54036 P(X = 1) = C(12,1)(0.05)(0.95)¹¹ = 12 x 0.05 x 0.56880 = 0.34128 P(X >= 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.54036 - 0.34128 = 0.11836 P(X = 2) = C(12,2)(0.05)²(0.95)¹⁰ = 66 x 0.0025 x 0.59874 = 0.09879 P(X = 3) = C(12,3)(0.05)³(0.95)⁹ = 220 x 0.000125 x 0.63025 = 0.01733 P(2 <= X <= 3) = 0.09879 + 0.01733 = 0.11612 Step 5: Divide. 0.11612 / 0.11836 = 0.9811 Required probability approx 0.981 (3 s.f.). (It is close to 1 because a rejected carton almost always has just 2 or 3 seasonal boxes; 4 or more is very unlikely at p = 0.05.)