Answer :
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 [tex]\frac{1}{2}[/tex].
The probability of the micro controller's success is [tex]\frac{1}{2}[/tex].
Hence, the required probability is [tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}[/tex].
(b)
From the three micro controller, one can be chosen in [tex]^3C_1 = 3[/tex] ways.
The probability here is [tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\times3 = \frac{3}{8}[/tex].
(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 [tex](\frac{1}{2} )^4 = \frac{1}{16}[/tex].
If the chosen one does not work properly, then the probability of other 3 will work properly is [tex](\frac{1}{2} )^4 = \frac{1}{16}[/tex].
The required probability is [tex]3(\frac{1}{16} + \frac{1}{16} ) = \frac{3}{8}[/tex].
(d)
In this case there are total 6 micro controllers.
From these 6 controllers, 3 can be chosen as [tex]^6C_3 = \frac{6!}{3!\times3!} = 20[/tex]ways.
The probability that the team is able to reshuffle the micro controllers to make one robot work is [tex]20\times(\frac{1}{2} )^6 = \frac{5}{16}[/tex].