Out of 10 computer chips, four are defective. Find the following. If three chips are randomly chosen for testing (without replacement), compute the probability that at most 2 of them are defective.
Three scenarios possible, zero chips are defective. One chip is defective and lastly two chips are defective. Using Combination Formula.
$$ C(n,k)=\frac{n!}{k!(n-k)!}. $$ defective $=4 $
non defective $= 6$
Total Ways =$\binom{10}{3}$
X= $2 $ defective chips from $3 $ selected chips.
$$P(X\le2)=P(0)+P(1)+P(2)\\\frac{\binom{4}{0}\binom{6}{3}+\binom{4}{1}\binom{6}{2}+\binom{4}{2}\binom{6}{1}}{\binom{10}{3}}=.96667$$
Based on the above, would this be an appropriate way to solve the problem?
As @lulu mentioned in the comment, it is easier to compute the complement. You can also solve an equivalent problem:
$3$ out of $10$ chips are used for test. If $4$ chips are randomly chosen to be defective, what is the probability that not all test chips are defective?
Once again, use complement: $1-\binom{7}{1}/\binom{10}{4}$