Probability

Solution set: { 6, (1,5), (1,6) ......}

                  i.e. P(6 appeared on first throw) +

                       P(1 appeared on first throw and 5 appeared on second throw) +

                        P(1 appeared on first throw and 6 appeared on second throw) + ....................

                      = 1/6 + (1/6)(1/6) + (1/6)(1/6) + .....

                      = 1/6 + 9/36

                      = 5/12.

Viewpoint 2:

P(......) =   P(6 came on first throw) +  P(sum>= 6 and 1,2,3 appeared in first throw)

          =      1/6                                +                 ????

P(1,2,3 appeared in first throw) = 1/2                                                   //P(E1)

P(sum >= 6 | 1,2,3 appeared in first throw) = 9/18                             //P(E2 | E1)

// Our new sample space is:  { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

                                                 (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

                                                (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) }

// 9 favorable cases: {(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) }

P(sum>= 6 and 1,2,3 appeared in first throw) = (1/2)(9/18)              //P(E2 ∩ E1) = P(E1)P(E2|E1)

P(what we are looking for) = 1/6 + 9/36     = 5/12

Correct Answer: B

A computer can be declared faulty in two cases 
1) It is actually faulty and correctly declared so (p*q)
2) Not faulty and incorrectly declared (1-p)*(1-q).