Question

It is known from past experience that in a certain plant there are on the average $4$ industrial accidents per month. Find the probability that there will be less than $4$ accidents in a given month. Given: $({\mathrm{e}}^{-4}=0.0183)$

• (A) $0.423$
• (B) $0.433$
• (C) $0.443$
• (D) $0.453$

Answer 1

The Correct Option is B

Explanation:
Given:$\mathrm{m}=4$The probability distribution is given by:$\mathrm{p}(\mathrm{x})=\frac{({\mathrm{e}}^{-\mathrm{m}}\times {\mathrm{m}}^{\mathrm{x}})}{\mathrm{x}!}$Calculation:According to the formula,$\Rightarrow \mathrm{p}(\mathrm{x})=\frac{({\mathrm{e}}^{-4}\times 4^{\mathrm{x}})}{\mathrm{x}!}$Now, the probability that in a given month there will be less than $4$ acc idents$=\mathrm{p}(0)+\mathrm{p}(1)+\mathrm{p}(2)+\mathrm{p}(3)$$=\frac{\sum _{x=0}^{x=3}{\mathrm{e}}^{-4}{4}^x}{\mathrm{x}!}$$={\mathrm{e}}^{-4}[(1+4+8)+\frac{32}{3}]$$={\mathrm{e}}^{-4}[(13)+\frac{32}{3}]$$=0.0183\times \frac{71}{3}$$=0.0183\times 23.666$$=0.433$$\therefore$ The probability that in a given month there will be less than $4$ accidents is $0.433$.Hence, the correct option is (B).

Answer 2

Given:
Average number of accidents per month, \(\lambda = 4\).
Using the Poisson distribution formula:
\(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)

For \(P(X \lt 4)\):
\(P(X \lt 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\)

Calculate each probability:
\(P(X = 0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4}\)

\(P(X = 1) = \frac{e^{-4} \cdot 4^1}{1!} = 4 \cdot e^{-4}\)

\(P(X = 2) = \frac{e^{-4} \cdot 4^2}{2!} = 8 \cdot e^{-4}\)

\(P(X = 3) = \frac{e^{-4} \cdot 4^3}{3!} = \frac{32 \cdot e^{-4}}{3}\)

Now sum these probabilities:
\(P(X \lt 4) = e^{-4} + 4 \cdot e^{-4} + 8 \cdot e^{-4} + \frac{32 \cdot e^{-4}}{3}\)

Calculate \(e^{-4}\):
\(e^{-4} \approx 0.0183\)
Now compute \(P(X \lt 4)\):
\(P(X \lt 4) = 0.0183 + 4 \cdot 0.0183 + 8 \cdot 0.0183 + \frac{32 \cdot 0.0183}{3}\)
\(P(X \lt 4) = 0.0183 + 0.0732 + 0.1464 + 0.1952\)
\(P(X \lt 4) = 0.4331\)

So, the correct option is (B) \(0.433\).