Answer to Question #224194 in Statistics and Probability for Reyad

Question

A lot consists of (6+a) articles, 3 with minor defects, and 3 with major defects. The ‘a’

articles are chosen at random from the lot without replacement. Compute the

probability by using basic probability concepts, both articles are good.

Answer 1

A lot consists of (6+a) articles, 6 with defects.

$a$ articles are chosen at random from the lot without replacement.

Suppose that $X$ be the number of articles with defects from $a$ articles chosen

$P(X=0)=\dfrac{\dbinom{a}{a}\dbinom{6}{0}}{\dbinom{6+a}{a}}=\dfrac{6!a!}{(6+a)!}$

If $2$ articles are chosen at random from the lot without replacement $(a\geq2).$

Suppose that $Y$ be the number of articles with defects from $2$ articles chosen

$P(Y=0)=\dfrac{\dbinom{a}{2}\dbinom{6}{0}}{\dbinom{6+a}{2}}=\dfrac{a!6!a!}{2!(a-2)!(6+a)!}$