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$20\%$ of the bolts produced in a factory are found to be defective. Find the probability that in a sample of $10$ bolts chosen at random exactly $2$ will be defective using ,Binomial distribution.

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  • A random variable $X$ is said to follow a binomial distribution of its probability mass function is given by
  • $P(X=x)=p(x)=\left\{\begin{array}{1 1}nC_xp^xq^{n-x},&x=0,1.......n\\0,&otherwise\end{array}\right.$
  • Constants of Binomial Distribution :
  • Mean = np
  • Variance = npq
  • Standard deviation =$\sqrt{ variance }=\sqrt{ npq}$
  • $X\sim N(\mu,\sigma)$
  • The parameters of the distribution are $n,p\quad X\sim B(n,p)$
Step 1:
Let $X$ be random variable denoting the no of defective bolts in a sample of 10 bolts chosen at random.
Probability that a bolt is defective=0.2
$\therefore \lambda=np$
$\Rightarrow 10\times .2=2$
Step 2:
Using a bionomial distribution :
$n=10,p=0.2\Rightarrow q=0.8$
$X\sim B(10,0.2)$
$P(X=x)=10C_x(0.2)^x(0.8)^{10-x}\qquad x=0,1,2.......10$
Step 3:
Probability of exactly 2 defective bolts
$\qquad\qquad=\large\frac{10\times 9}{1\times 2}\frac{1}{5^2}\frac{4^8}{5^8}$
$\qquad\qquad=\large\frac{45}{5^{10}}$$\times 2^{16}$
answered Sep 18, 2013 by sreemathi.v

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