A car firm has 2 cars, which are hired out day-by-day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that some demand is refused on a given day. [Given \(e^{-1.5}=0.2231\)]

Q.5 (A) (iii) A car firm has 2 cars, which are hired out day-by-day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that some demand is refused on a given day. [Given \(e^{-1.5}=0.2231\)]

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: 0.1913

Let \(X\) = number of cars demanded, \(X\sim P(m=1.5)\), \(P(X=x)=\frac{e^{-1.5}(1.5)^x}{x!}\).

Demand is refused when more than 2 cars are demanded:

\(P(X\gt 2)=1-[P(0)+P(1)+P(2)]=1-e^{-1.5}\left[1+1.5+\frac{(1.5)^2}{2}\right]\)

\(=1-0.2231\times 3.625=1-0.8087=0.1913\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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A car firm has 2 cars, which are hired out day-by-day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that some demand is refused on a given day. [Given \(e^{-1.5}=0.2231\)]

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