Three students S_1, S_2 and S_1 are given a problem to solve. Consider the following events : U: At least one of S_1, S_2 and S_3 can solve the problem, V: S_1 can solve the problem, given that neither S _2 nor S _3 can solve the problem, W: S_2 can solve the problem and S_3 cannot solve the problem, T: S_3 can solve the problem. For any event E, let P(E) denote the probability of E. If P(U)= 1 2 , P(V)= 1 10 and P(W)= 1 12 , then P(T) is equal to
A. 13 36
aligned & P ( U )=1- P ( ~S _1^ S _2^ S _3^ )= 1 2 \\ & P ( ~S _1^ S _2^ S _3^ )= 1 2 ; P ( ~S _1^ ) P ( ~S _2^ ) P ( ~S _3^ )= 1 2 \\ & (1- P ( ~S _1 ) ) (1- P ( ~S _2 ) ) (1- P ( ~S _3 ) )= 1 2 (1) aligned aligned & P ( V )= P ( S _1 ~S _2^ S _3^ ) P ( S _2^ S _3^ ) = 1 10 \\ & P ( S _1 ) P ( S _2^ ) P ( S _3^ )= 1 10 P ( S _2^ ) P ( S _3^ ) \\ & P ( S _1 )= 1 10 aligned aligned & P ( ~W )= P ( ~S _2 ~S _3^ )= 1 12 \\ & P ( ~S _2 ) P ( ~S _3^ )= 1 12 \\ & P ( ~S _2 ) (1- P ( ~S _3 ) )= 1 12 ...(2) aligned Eq. (1) (1- 1 10 ) (1- P ( ~S _2 ) ) (1- P ( ~S _3 ) )= 1 2 (1- P ( ~S _2 ) ) (1- P ( ~S _3 ) )= 5 9 ....(3) aligned & Eq. (2) Eq. (3) P ( ~S _2 ) 1- P ( ~S _2 ) = 1 12 9 5 \\ & P ( ~S _2 )= 3 23 aligned Put in Eq. (2) aligned & 3 23 (1- P ( ~S _3 ) )= 1 12 \\ & 1- P ( ~S _3 )= 23 36 \\ & P ( ~S _3 )= 13 36 \\ & P ( ~T )= 13 36 aligned
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