Question

If $10$ pens are drawn successively with replacement from a box of pens in which $10\%$ pens are defective. The probability that at least $1$ pen is defective is

• A. $1-{\left(\frac{9}{10}\right)}^{10}$
• B. $1-{\left(\frac{1}{10}\right)}^{10}$
• C. $\left(\frac{9}{10}\right)$
• D. ${\left(\frac{9}{10}\right)}^9$

Answer 1

The correct option is A $1-{\left(\frac{9}{10}\right)}^{10}$

P(at least one pen is defective)$=1-$P(no pen is defective)
$=1{-}^{10}{C}_0(\frac{9}{10}{)}^{10}=1-(\frac{9}{10}{)}^{10}$