probability that at least two part will be defective ## Question > The probability that a part manufactured by a company will be defective is $0.05$. If $15$ such parts are selected randomly and inspected, then the probability that at least two part will be defective is ________. (round off to two decimal places) ## My Approach $$P(\text{atleast 2 part defective})=1-P(\text{No part defective}) -P(\text{1 part defective})$$ $$=1-(0.95)^{15}-((0.95)^{14} \times 0.05)$$ Is it correct?

Not quite, $$P=1-(0.95)^{15}-{15\choose 1}((0.95)^{14} \times 0.05)$$