Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# There are two lots of identical articles with different amounts of standard and defective articles. There are $N$ articles 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 of results. Find the probability that an article selected at random from the new lot is defective.a) $\dfrac{\text{KnM+LmN}}{MN(K+L)}$b) $\dfrac{\text{KnM+MmN}}{MN(K+M)}$c) $\dfrac{\text{LnM+LmN}}{MN(K+M)}$d) $\text{1-}\left( \dfrac{\text{KnM+LmN}}{MN(K+L)} \right)$

Last updated date: 17th Jun 2024
Total views: 413.7k
Views today: 8.13k
Verified
413.7k+ views
Hint: Use the concept of basic definition of probability and the rule of AND and OR between the events.
$P(Event)=\dfrac{\rm{Favourable \space cases}}{\rm{Total \space cases}}$
$P\left( A\text{ }or\text{ }B \right)\text{ }=\text{ }P\left( A \right)\text{ }+\text{ }P\left( B \right)$
$P\left( A\text{ }and\text{ }B \right)\text{ }=\text{ }P\left( A \right)\text{ }\times \text{ }P\left( B \right)$
We need to find the probability of an article selected at random from the new lot being defective.

Given: There are two lots of identical articles with different amounts of standard and defective articles.
$N$ article in the first lot, $n$ of which are defective
$M$ 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 of results.

$P({{E}_{1}})$: The selected article is from ${{1}^{st}}$ lot $=\dfrac{\rm{Number\text{ }of\text{ }articles\text{ }selected\text{ }from\text{ }the\text{ }first\text{ } lot}}{\rm{Total\text{ }article\operatorname{s}\text{ }in\text{ }the\text{ }new\text{ } lot\text{ }formed}}$ $=\dfrac{K}{K+L}$

$P({{E}_{2}})$ : The selected article is from ${{2}^{nd}}$ lot $=\dfrac{\rm{Number\text{ }of\text{ }articles\text{ }selected\text{ }from\text{ }the\text{ }\sec ond\text{ } lot}}{\rm{Total\text{ }article\operatorname{s}\text{ }in\text{ }the\text{ }new\text{ } lot\text{ }formed}}$ $=\dfrac{L}{K+L}$
Required Probability:\begin{align} & ={\rm(Particle\text{ }selected\text{ }at\text{ }random\text{ }from\text{ }the\text{ }new\text{ } lot\text{ }being\text{ }defective)} \\ & ={\rm(Particle\text{ }selected\text{ }from\text{ }the\text{ }first\text{ } lot\text{ }being\text{ }defective)\text{ }} or \text{ } {\rm(Particle\text{ }selected\text{ }from\text{ }the\text{ }\sec ond\text{ } lot\text{ }being\text{ }defective)} \\ \end{align}
\begin{align} & =\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{1}^{st}}}} lot)\text{ }} {\rm and} \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{1}^{st}}}} lot)}+\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{2}^{nd}}}} lot)\text{ }} {\rm and}\operatorname{\rm{P}(defective\text{ }articl{{e}_{{{2}^{nd}}}} lot)} \\ & =\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{1}^{st}}}} lot)}\times \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{1}^{st}}}} lot)}+\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{2}^{nd}}}} lot)}\times \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{2}^{nd}}}} lot)} \\ & =\left( \dfrac{K}{K+L} \right)\times \dfrac{n}{N}+\left( \dfrac{L}{K+L} \right)\times \dfrac{m}{M} \\ & =\dfrac{1}{K+L}\left( \dfrac{Kn}{N}+\dfrac{Lm}{M} \right) \\ & =\dfrac{LmN+KnM}{NM(K+L)} \\ \end{align}

Hence the correct answer is Option A.

Note: In such type of questions which involves probability knowing the definition of probability combined with OR and AND rule is needed. Accordingly follow the steps to find the required answer.