# If the probability of a defective fuse from a manufactuning unit is $2\%$ in a box of $200$ fuses find the probability that exactly $4$ fuses are defective

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 exactly 4 fuses are defective
$P(X=4)=\large\frac{e^{\Large -4}{4^4}}{4!}$
$\qquad\qquad=0.0183\times \large\frac{32}{3}$
$\qquad\qquad=0.0061\times 32$
$\qquad\qquad=0.1952$
edited Sep 18, 2013