1
$\begingroup$

4 out of 18 discs are defective , if 3 discs are selected determine the ratio of the number of ways of selecting at least 1 defective disc to the number of ways of selecting no defective discs at all.

Please help i am only able to get 765 : 91

The answer is 113 : 91

$\endgroup$
1
$\begingroup$

The number of ways of selecting no defective discs is $C(14,3) = 364$.

The number of ways of selecting any three discs is $C(18,3) = 816$.

So the number of ways of selecting at least one defective disc is $816-364 = 452$ (using the idea of a complement) and so the requested ratio is $452:364 = 113:91$.

Note that the phrase "at least one" means "not none" and this is an indication to count a complement.

$\endgroup$
0
$\begingroup$

The number of ways to select no defective disc at all is 18 choose 3. Selecting at least one defective disc is the complementary case of selecting no defective disc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.