Ask Question
19 June, 07:43

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?

+4
Answers (1)
  1. 19 June, 08:09
    0
    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) ]
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers