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# If the probability of a defective fuse from a manufactuning unit is $2\%$ in a box of $200$ fuses find the probability that more than $3$ fuses are defective$[e^{-4}=0.0183].$

Toolbox:
• A random variable $X$ is said to have a poisson distribution of the probability mass function of $X$ is
• $P(X=x)=\large\frac{e^{\Large -\lambda }\lambda^x}{x!}$$\qquad (x=0,1,2........ for some \lambda > 0) • Constants of a poisson distribution : • Mean=Variance=\lambda • A continuous random variable X is said to follow a normal distribution with parameter \mu and \sigma (or \mu and \sigma^2) if the probability density function is • f(x)=\large\frac{1}{\sigma \sqrt{2\pi}}$$e^{-\large\frac{1}{2}(\frac{x-\mu}{\sigma})^2};-\infty < x < \infty,-\infty< \mu <0$ and $\sigma > 0$
• $X\sim N(\mu,\sigma)$
• Constants of a normal distribution :
• Mean =$\mu$,variance =$\sigma^2$,standard deviation =$\sigma$
Step 1:
Let $X$ be the random variable denoting the number of defective fuses in a box of 200 fuses.
Probability of a defective fuse P=0.02
$\lambda=np$
$\quad=200\times 0.02$
$\quad=4$
$\therefore X\sim P(4)$
$\Rightarrow P(X=x)=\large\frac{e^{-4}4^4}{4!}$$\qquad x=0,1,2......$
Step 2:
Probability that more than 3 fuses are defective.
$P(X > 3)=1-P(X \leq 3)=1-e^{-4}[\large\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}]$
$\qquad\qquad=1-0.0183[\large\frac{(1+4+8)^3+32}{3}]$
$\qquad\qquad=1-0.4331$
$\qquad\qquad=0.5669$