1. A machine comprises of three transformers A, B and C. Such machine may operate if at least 2 transformers are working. The probability of each transformer working is given as displayed below;
P (A) = 0.6, P (B) = 0.5, P (C) = 0.7
A mechanical engineer went to inspect the working situations of those transformers. Determine the probabilities of having the given outcomes
i. Only one transformer operating
ii. Two transformers are operating
iii. All three transformers are operating
iv. None is operating
v. At least two are operating
vi. At most two are operating
Solution
P(A) =0.6 P(A) = 0.4 P(B) = 0.5 P(~B) = 0.5
P(C) = 0.7 P(C¯) = 0.3
i. P(only one transformer is operating) is described by the given possibilities
1st 2nd 3rd
P (A B¯ C¯ ) = 0.6 x 0.5 x 0.3 = 0.09
P (A B C¯ ) = 0.4 x 0.5 x 0.3 = 0.06
P (A B¯ C) = 0.4 x 0.5 x 0.7 = 0.14
∴ P(Only one transformer working)
= 0.09 + 0.06 + 0.14 = 0.29
ii. P(only two transformers are operating) is described by the given possibilities.
1st 2nd 3rd
P (A B C¯ ) = 0.6 x 0.5 x 0.3 = 0.09
P (A B¯ C) = 0.6 x 0.5 x 0.7 = 0.21
P (A B C) = 0.4 x 0.5 x 0.7 = 0.14
∴ P(Only two transformers are operating)
= 0.09 + 0.21 + 0.14 = 0.44
iii. P(all the three transformers are operating).
= P(A) x P(B) x P(C)
= 0.6 x 0.5 x 0.7
= 0.21
iv. P(none of the transformers is operating).
= P(A) x P(B¯) x P(C¯)
= 0.4 x 0.5 x 0.3
= 0.06
v. P(at least 2 working).
= P(exactly 2 working) + P(all three working)
= 0.44 + 0.21
= 0.65
vi. P(at most 2 working).
= P(Zero working) + P(one working) + P(two working)
= 0.06 + 0.29 + 0.44
= 0.79