Question

There are two lots of identical articles with different amount of standard and defective articles. There are N article in the first lot, n of which are defective and M articles in the second lot, m of which are defective. K articles are selected from the first lot and L articles from the second and a new lot results. Find the probability that an article selected at random from the new lot is defective

• A. $\frac{KnM+LmN}{MN(K+L)}$
• B. $\frac{KnM+MmN}{MN(K+M)}$
• C. $\frac{LnM+LmN}{MN(K+M)}$
• D. $1-\frac{KnM+LmN}{MN(K+L)}$

Answer 1

The correct option is A $\frac{KnM+LmN}{MN(K+L)}$

$P\left(E_1\right)$ : The selected article is from $I^{st}$ lot $=\frac{K}{K+L}$
$P\left(E_2\right)$ : The selected article is from $I{I}^{nd}$ lot $=\frac{L}{K+L}$
Required probability $=\frac{K}{K+L}.\frac{n}{N}+\frac{L}{K+L}.\frac{m}{M}=\frac{LMn+LmN}{NM(K+L)}$