In a batch of 10 laser diodes, two have efficiency below 0.28, six have efficien
ID: 3300858 • Letter: I
Question
In a batch of 10 laser diodes, two have efficiency below 0.28, six have efficiency between 0.28 and 0.35, and two have efficiency above 0.35. Two diodes are selected at random and without replacement. Are the events E_1 = {the first diode selected has efficiency below 0.28} and E_2 = {the second diode selected has efficiency above 0.35} independent? Justify your answer. The system of four components shown in figure below functions as long as components 1 and 2 both function or components 3 and 4 both function. Each of the four components functions with probability 0.9 independently of the others. Find the probability that the system functions.Explanation / Answer
5) P(E1) =2/10=1/5 ( as 2 have efficiency below 0.28 out of 10 )
P(E2) =P( first diode has effciiency above 0.35 and second has efficiency above 0.35+first diode does not have efficiency above 0.35 and second has efficiency above 0.35)
=(2/10)*(1/9)+(8/10)*(2/9) =18/90 =1/5
and Probability P(E1nE2) P( first diode below 0.28 and second above 0.35) =(2/10)*(2/9) =2/45
as P(E1nE2) is not equal to P(E1)*P(E2) ; therefore E1 and E2 are not independent.
(Note: we can see that E2 depends on which type of diode is drawn on first draw)
6) probability that 1 and 2 both functions =P(A) =0.9*0.9=0.81
robability that 3 and 4 both functions =P(B) =0.9*0.9=0.81
probability that system functions =P( atleast one of A or B works) =1-P( none of A and B works)
=1-(1-P(A))*(1-P(B)) =1-(1-0.81)*(1-0.81)=0.9639
please revert for any clarification required