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?

Answer 1

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 $^6C_3 = \frac{6!}{3!\times3!} = 20$ways.

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