Three microcontrollers are needed to operate a specific type of robot. Such a robot stops working whenever one or more microcontrollers fail

Question

Three microcontrollers are needed to operate a specific type of robot. Such a robot stops working whenever one or more microcontrollers fail. On competition day, the probability

that a microcontroller fails is equal to (1/2), independently of other microcontrollers.

(a) 1 pt – What is the probability that a robot with 3 microcontrollers works on competition day?

(b) 1 pt – What is the probability that exactly one microcontroller has failed given that the robot is not working?

(c) 1.5 pt – If the team starts the day with four microcontrollers and one robot, what is the probability that they can find a suitable configuration for the robot to work on competition day?

(d) 1.5 pt – Suppose that 2 robots, each with 3 microcontrollers, are brought to a competition site. Upon initial testing, both robots are not working. What is the probability that the team is able to reshuffle the microcontrollers to make one robot work?

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Sarah 2 weeks 2021-10-10T00:02:30+00:00 1 Answer 0

Answers ( )

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    2021-10-10T00:04:26+00:00

    Answer:

    See the explanation.

    Step-by-step explanation:

    (a)

    The robot will work only if all the micro controller works on the competition day.

    The probability of the micro controller’s failing is \frac{1}{2}.

    The probability of the micro controller’s success is \frac{1}{2}.

    Hence, the required probability is \frac{1}{2} \times \frac{1}{2}  \times \frac{1}{2} = \frac{1}{8}.

    (b)

    From the three micro controller, one can be chosen in ^3C_1 = 3 ways.

    The probability here is \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\times3 = \frac{3}{8}.

    (c)

    If from the four micro controllers, one fails, then also they can manage to make  the robot work.

    From the 4, 1 can be chosen in 4 ways.

    This one can either work properly or not.

    If it works properly, then the probability of other 3 will work properly is (\frac{1}{2} )^4 = \frac{1}{16}.

    If the chosen one does not work properly, then the probability  of other 3 will work properly is (\frac{1}{2} )^4 = \frac{1}{16}.

    The required probability is 3(\frac{1}{16} + \frac{1}{16}  ) = \frac{3}{8}.

    (d)

    In this case there are total 6 micro controllers.

    From these 6 controllers, 3 can be chosen as ^6C_3 = \frac{6!}{3!\times3!} = 20ways.

    The probability that the team is able to reshuffle the micro controllers to make one robot work is 20\times(\frac{1}{2} )^6 = \frac{5}{16}.

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