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Q)A product is an assembly of 4 different components. The product can be sequentially assembled in two possible ways. If the 4 components are placed in a box and these are drawn at random from the box, then the probability of getting a correct sequence is ______.
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Assume we have $4$ different components as : $\{P_1,P_2,P_3,P_4\}$

These components are sequentially assembled in two different ways implies that all permutation of these components would not give the desire result.
So bind two components : $\{(P_1,P_2),(P_3,P_4)\}$

Working sequences  can be find as : $\large\frac{4!}{2!\times2!}=6$

which are as follows :

$\{P_1,P_2,P_3,P_4\}$

$\{P_1,P_3,P_2,P_4\}$

$\{P_1,P_3,P_4,P_2\}$

$\{P_3,P_1,P_2,P_4\}$

$\{P_3,P_1,P_4,P_2\}$

$\{P_3,P_4,P_1,P_2\}$

Idea is much more similiar to serializability concept , here I assumed that $P_2$ cannot be executed before $P_1$ and $P_4$ before $P_3.$

Now,

Probability of getting working sequence : $\Large{\frac{6}{4!}} \ =\frac{6}{24}=\frac{1}{4}$

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