Question3 In the circuit below, each switch may be open or closed randomly. The

Question

Question3 In the circuit below, each switch may be open or closed randomly. The event that a particular switch is closed has probability given by the number next to the switch in the circuit diagram, and these five events are independent. a) What is the probability that a connection exists between X and Y? b) What is the probability that switch A is open given that a connection exists between X and Y? c) What is the probability that switch A is open given that a connection does not exist between X and Y? 0.7 0.7 0.8 0.8 08

Explanation / Answer

(a)

P(connection exists) = 1 - P(connection does not exist)

P(connection does not exist) = P(upper wire is open)*P(lower wire is open)

Now,

P(upper wire is open) = P(A)*P(B') + P(A')*P(B) = 0.7*0.3 + 0.3*0.7 = 0.42

P(lower wire is open) = 1 - P(lower wire is closed) = 1 - (0.8*0.8*0.8) = 0.488

So,

P(connection does not exist) = 0.42*0.488 = 0.205

So,

P(connection exists) = 1 - 0.205 = 0.795

(b)

P(switch A is open | connection exists) = P(switch A is open AND connection exists)/P(connection exists)

P(switch A is open AND connection exists) = 0.3*0.7*0.8*0.8*0.8 + 0.3*0.3*0.8*0.8*0.8 = 0.1536

So,

P(switch A is open | connection exists) = 0.1536/0.795 = 0.193

(c)

P(switch A is open | connection does not exist) = P(switch A is open AND connection does not exist)/P(connection does not exist)

P(switch A is open AND connection does not exist) = P(A')*P(B)*P(lower wire is open) + P(A')*P(B')*P(lower wire is open) = 0.3*0.7*0.488 + 0.3*0.3*0.488 = 0.1464

So,

P(switch A is open | connection does not exist) = 0.1464/0.205 = 0.714

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