Question

A company uses a combination of three components- A, B and C to create three different drone designs. The first design Glider uses 3 parts of component A and 2 parts of components B. Design Blimp uses 2 parts of component B and C, and the last design, Pilot uses one part of each component. A sample of 75 components, 25 A, 25 B, 25 C, will be used to make prototypes for the various designs. If 30 components are selected at random, what is the likelihood two prototypes of each design can be made?

Answer 1

Total ways of selection are

The problem first wants us to formulate the minimum number of components of each of the three types, that will be required to design two prototypes of each type. The below table summarizes the required number of components

	Glider	Blimp	Pilot	Total
A	3		1	4
B	2	2	1	5
C		2	1	3

Hence, if we need at least one prototype of each design, then we need at least A = 4, B = 5 and C = 3. So for two prototypes, we need A = 8, B = 10 and C = 6, viz. a total of 24 out of the selection of 30 from 75 components, 25 of each type.

So the problem statement now becomes the likelihood of selecting at least 8 of component A, 10 of component B and 6 of component C when a total of 30 are chosen from a sample space of 25 of each of the three components.

So 6 components can be chosen freely while 24 must be chosen as per the constraint. Now we should find the distinct ways of distributing the 6 components that are to be chosen from the three types.

All possible pairs (a, b, c) such that a + b + c = 6 are

(a, b, c) = (6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 2, 2), (4, 1, 1), (1, 4, 1), (1, 1, 4), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)

We need to add (a₀, b₀, c₀) = (8, 10, 6) to each of the above triplets, that are favorable selections. These are 13 different possible selections satisfying the given constraint. For short notation, if we define

then the required favorable ways of selection are

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