IB Mathematics AHL 4.17 Poisson distribution -AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed solution
(a)
Let \(X \sim Po(1.2)\).
\(P(X=1) = \frac{e^{-1.2}(1.2)^1}{1!} \approx 0.361\).
(b)
\(P(X \ge 1) = 1 – P(X=0)\)
\(P(X \ge 1) = 1 – e^{-1.2} \approx 0.699\).
(c)
For \(5\) quadrats, the sum of independent Poisson variables is also Poisson with \(\lambda_{total} = 5 \times 1.2 = 6\).
Let \(Y \sim Po(6)\).
\(P(Y=5) = \frac{e^{-6}(6)^5}{5!} \approx 0.161\).
(d)
This requires exactly one worm in each of the 5 quadrats independently.
Probability \(= (P(X=1))^5\)
\(= (0.3614…)^5 \approx 0.00617\).
(e)
Let \(W\) be the number of quadrats with at least one worm.
\(W \sim B(5, p)\) where \(p = P(X \ge 1) \approx 0.699\) (from part b).
We need \(P(W=3)\).
\(P(W=3) = \binom{5}{3} (0.6988)^3 (1-0.6988)^2 \approx 0.310\).