A valve can be used at four temperature levels If the valve

A valve can be used at four temperature levels. If the valve is used at the cold temperature, then there is a probability of 0.003 that it will leak. If the valve is used at the medium temperature, then there is a probability of 0.009 that it will leak. If the valve is used at a warm temperature, then there is a probability of 0.018 that it will leak. If the valve is used at hot temperature, then there is a probability of 0.014 that it will leak. Under standard operating conditions, the valve is used at a clod temperature 12% of the time, at a medium temperature 55% of the time, at warm temperature 20% of the time, and at hot temperature 13% of the time.

A) If the value leaks, what is the probability that it is being used at the hot temperature?

B) If the valve does not leak, what is the probability that it is being used at the medium temperature?

Solution

Under standard opearting conditions;

Let:

P(C)= Probabilty that the valve is used at cold temperatures

P(M)=Probabilty that the valve is used at medium temperatures

P(W)=Probabilty that the valve is used at warm temperatures

P(H)=Probabilty that the valve is used at hot temperatures

P(L/C)= Probability that the valve leaks given that it is used at cold temperatures

P(L/M)=Probability that the valve leaks given that it is used at medium temperatures

P(L/W)=Probability that the valve leaks given that it is used at warm temperatures

P(L/H)=Probability that the valve leaks given that it is used at hot temperatures

We know from the question that:

P(C)=0.12, P(M)= 0.55, P(W)= 0.20, P(H)= 0.13 and

P(L/C)=0.003, P(L/M)= 0.009, P(L/W)=0.018, P(L/H)=0.014

The total probability that the valve leaks is P(L) where:

P(L)=P(C)XP(L/C)+ P(M)XP(L/M)+P(W)XP(L/W)+P(H)XP(L/H)=0.12X0.003+0.55X0.009+0.2X0.018+0.13X0.014= 0.044

Now, P(H/L)= Probability that the valve leaks and is being used at hot temperatures given that it leaks

P(H/L)= P(H)XP(L/H) / P(L)

           =0.00182 / 0. 0.01073

           = 0. 17036

           =17.04%

If the value leaks, the probability that it is being used at the hot temperature is 17.04%

PART B

P(NL)= probability that the valve does not leak:

P(NL/C): Probability that the valve doe not leak given that it is used at cold temperature=0.997(1-0.003)

P(NL/M):Probability that the valve doe not leak given that it is used at medium temperature=0.991(1-0.009)

P(NL/W):Probability that the valve doe not leak given that it is used at warm temperature=0.982(1-0.018)

P(NL/H):Probability that the valve doe not leak given that it is used at hot temperature=0.986(1-0.014)

P(NL)= P(C)XP(NL/C) + P(M)XP(NL/M) + P(W)XP(NL/W) + P(H)XP(NL/H)

         =0.12X0.997 +0.55X0.991 + 0.20X0.982 + 0.13X0.986

         =0.98927

P(M/NL): probability that the valve is used at medium temperature and does not leak given that it does not leak

P(M/NL)= P(M)XP(NL/M) / P(NL)

             =0.55X 0.991 / 0.98927

             =0.54505 / 0.98927

              = 0.55096

              = 55.096%

If the valve does not leak, the probability that it is being used at the medium temperature is 55.096%