Examples on probability, Mathematics

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

Posted Date: 2/20/2013 4:31:15 AM | Location : United States







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