F5 probability(急)

7(a)(i) P(both cables X and Z fail)
= 0.015*0.03
= 0.00045(ii) P(both cables X, Y and Z fail)
= 0.015*0.025*0.03
= 0.00001125(iii)
P(A and B will be able to make contact)
= P(X and Y are operative) + P(Z is operative) - P(both cables X, Y and Z are operative)
= (0.985)(0.975) + 0.97 - (0.985)(0.975)(0.97)
= 0.99881125P(A and B will not be able to make contact)
= 1 - 0.99881125
= 0.00118875(b) P(A and B will not be able to make contact | X fails)
= P(Z fails)
= 0.03(c) P(X fails | A and B are not able to contact)
= (0.015)(0.03)/0.00118875
= 0.3785

(b) Given that X fails to operate.

a i) P(X' ∩ Z') = 0.015 x 0.03 = 0.00045

ii) P(X' ∩ Y' ∩ Z') = 0.015 x 0.025 x 0.03 = 0.00001125

iii) P(A and B not able to make contact) = P[Z' ∩ (X' ∪ Y')]

Now P(X' ∪ Y') = P(X') + P(Y') - P(X' ∩ Y') = 0.025 + 0.015 - 0.025 x 0.015 = 0.03693

Hence P[Z' ∩ (X' ∪ Y')] = P(Z') x P(X' ∪ Y') = 0.03 x 0.03693 = 0.001189

b) Indeed it is P(X' ∪ Y') = 0.03693

c) P(X')/P(X' ∪ Y') = 0.3785