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?

Question

Mathematics

Teagan Davidson

8 December, 20:29

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?

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Answers (1)

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Aleena Day · Answer 1

Consider the following calculations

Step-by-step explanation:

Since 1 Blimp uses 2 components of B and C each

=> choosing 2 components of B (remaining after using in other prototypes) for 1st model = 22C2

choosing 2 components of B (remaining after using in other prototypes) for 2nd model = 21C2

choosing 2 components of B (remaining after using in other prototypes) for 3rd model = 20C2

choosing 2 components of B (remaining after using in other prototypes) for 4th model = 19C2

choosing 2 components of B (remaining after using in other prototypes) for 5th model = 18C2

and choosing 2 components of C (remaining after using in other prototypes) = 24C2

Similarly for C

P (5 prototypes of Blimp created) = [ (22C2 / 25C2) * (24C2 / 25C2) ] + [ (21C2 / 25C2) * (23C2 / 25C2) ]+[ (20C2 / 25C2) * (22C2 / 25C2) ]+[ (19C2 / 25C2) * (21C2 / 25C2) ]+[ (18C2 / 25C2) * (20C2 / 25C2) ]