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📅 March 10, 2026 👤 scraface

Mar 10, 2026

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Use binomial distribution: \( n = 150 \), \( p = 0.03 \), \( k = 5 \).
\( P(X=5) = \binom{150}{5} (0.03)^5 (0.97)^{145} \).
Approximate with Poisson: \( \lambda = 150 \times 0.03 = 4.5 \).
\( P(X=5) = \frac{e^{-4.5} (4.5)^5}{5!} \).
\( e^{-4.5} \approx 0.011109 \), \( 4.5^5 = 1845.5625 \), \( 5! = 120...
\( P = \frac{0.011109 \times 1845.5625}{120} \approx \frac{20.48}{120}...
More accurately: \( P = \binom{150}{5} (0.03)^5 (0.97)^{145} \).
\( \binom{150}{5} = \frac{150 \cdot 149 \cdot 148 \cdot 147 \cdot 146}...
\( (0.03)^5 = 2.43 \times 10^{-8} \)
\( (0.97)^{145} = e

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