Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

Let X denote number of defective bolts. A bolt can either be defective or not defective. So it is problem of bionomial distribution

Let P be probability a bolt is defective = 10% = 0.1

$$q = 1-p = 1-0.1 = 0.9$$

$$\mathrm{\therefore } n = 5$$

For bionomial distribution

$$P(X=x) = \mathrm{\ }n_x\ p^x\ q^{n-x}=\ \mathrm{\ }5_{C_x}\ {0.1}^x\ {0.9}^{5-x}$$

P(Atmost 1 will be defective ) $$P(X=0) + P(X=1)$$ $$= 5_{C_0}\ {0.1}^0\ {0.9}^{5-0}+5_{C_1}\ {0.1}^1\ {0.9}^{5-1}$$ $$= 0.5905 + 0.3281$$

$$= 0.9186$$

$\boldsymbol{\mathrm{\therefore }}$ Hence P atmost one will be defective = 0.9186