Respuesta :
Answer:
(a) 0.970225.
(b) 0.999775.
(c) 2.25 x [tex]10^{-4}[/tex].
Step-by-step explanation:
Given the probability that either system will function satisfactorily during a flight is 0.985.
(a) To calculate the probability that both systems function satisfactorily we are given the probability of either system will function of 0.985 which means that both function have probability of functioning satisfactorily of 0.985 each i.e.,
= 0.985 x 0.985 = 0.970225.
(b) The probability of first function will not function satisfactorily is 1 - 0.985 because the probability of first system function satisfactorily is 0.985.
Similarly, probability of second function will not function satisfactorily is also 1 - 0.985 = 0.015
So Probability that during a given flight at least one system functions satisfactorily = 1 - both system will not function satisfactorily
= 1 - (0.015 x 0.015) = 0.999775.
(c) Probability that during a given flight both systems fail or not function satisfactorily = (1 - 0.985) x (1 - 0.985) = 2.25 x [tex]10^{-4}[/tex]
Answer:
(a) P (Both systems functioning satisfactorily) = 0.970
(b) P (At least one system functions satisfactorily) = 0.999
(c) P (Both the systems failing) = 0.00023
Step-by-step explanation:
Let A = the system in use is working satisfactorily and B = the backup system is working satisfactorily.
The probability that either of the systems works satisfactorily is,
P (A) = P(B) = 0.985
(a)
Both the events A and B are independent, i.e. [tex]P(A\cap B)=P(A)\times P(B)[/tex]
Compute the probability that during a given flight both systems function satisfactorily as follows:
[tex]P(A\cap B)=P(A)\times P(B)[/tex]
[tex]=0.985\times0.985\\=0.970225\\\approx0.970[/tex]
Thus, the probability of both systems functioning satisfactorily is 0.970.
(b)
Compute the probability that during a given flight at least one system functions satisfactorily as follows:
P (At least one system functions satisfactorily) = 1 - P (None of the system functions satisfactorily)
= [tex]1-[P(A^{c})\times P(B^{c})][/tex]
[tex]=1-([1-P(A)]\times [1-P(B)])\\=1-([1-0.985]\times [1-0.985])\\= 1-0.000225\\=0.999775\\\approx0.999[/tex]
Thus, the probability that during a given flight at least one system functions satisfactorily is 0.999.
(c)
Compute the probability that during a given flight both the systems fail as follows:
P (Both the systems failing) = [tex]P(A^{c})\times P(B^{c})[/tex]
[tex]=[1-P(A^{c})]\times [1-P(B^{c})]\\=(1-0.985)\times (1-0.985)\\=0.015\times 0.015\\=0.000225\\\approx0.00023[/tex]
Thus, the probability that during a given flight both the systems fail is 0.00023.
