Question

The items produced by a firm are supposed to be $5$% defective what is the probability that a sample of $8$ items will contain less than $2$ defective items?

Answer 1

X=number of defective items

p= probability of getting a defective item = 5%=5/100

q= 1-p = 1-(5/100)=95/100

P(X< 2)

=P(X=0)+P(X=1)

$={}^8{C}_0\left(\frac{5}{100}{)}^0\right(\frac{95}{100}{)}^8+{}^8{C}_1\left(\frac{5}{100}{)}^1\right(\frac{95}{100}{)}^7+.......$

$=(\frac{95}{100}{)}^8+8\cdot (\frac{5\cdot {95}^7}{{100}^8})$

$=\left(\frac{{95}^7}{{100}^8}\right)(95+40)$

$=135\cdot \left(\frac{{95}^7}{{100}^8}\right)$

$=135\cdot \left(\frac{19}{20}{)}^7\right(\frac{1}{100})$

$=\left(\frac{27}{20}\right)\cdot (\frac{19}{20}{)}^7$

$=\left(\frac{27\cdot {19}^7}{{20}^8}\right)$