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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 exactly $4$ fuses are defective

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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
$P(X=4)=\large\frac{e^{\Large -4}{4^4}}{4!}$
$\qquad\qquad=0.0183\times \large\frac{32}{3}$
$\qquad\qquad=0.0061\times 32$
$\qquad\qquad=0.1952$
answered Sep 18, 2013 by sreemathi.v
edited Sep 18, 2013 by sreemathi.v
 

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