Question

10 bulbs out of a sample of 100 bulbs manufactured by a company are defective. The probability that 3 out of 4 bulbs, bought by a customer will not be defective, is

• A. $\frac{{}^4{C}_3}{{}^{100}{C}_4}$
• B. $\frac{{}^{90}{C}_3}{{}^{96}{C}_4}$
• C. $\frac{{}^{90}{C}_3}{{}^{100}{C}_4}$
• D. $\frac{{(}^{90}{C}_3{\times }^{10}{C}_1)}{{}^{100}{C}_4}$

Answer 1

The correct option is D $\frac{{(}^{90}{C}_3{\times }^{10}{C}_1)}{{}^{100}{C}_4}$

100 bulbs = 10 defective + 90 non defective probability that 3 out of 4 bulbs, bought by a customer will not to
be defective.
$=\frac{{}^{90}{C}_3{\times }^{10}{C}_1}{{}^{100}{C}_4}$