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Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is at least one closed path when the relays are activated. The relays may malfunction and not close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of .9. Given that current flowed when the relays were activated, what is the probability that relay 1 functioned?

Respuesta :

Answer:

0.9009

Step-by-step explanation:

Let X be the event that Current will flow

Let Y be the event that the first relay which is 1 is closed

Thus, we can say that every element of Y is in X, but X possesses more elements. Thus, Y ⊂ X.

Thus, we can say that;

P(X ∩ Y) = P(Y)

Thus, given that current flowed, the probability that relay 1 functioned will be expressed as;

P(Y | X) = (P(Y ∩ X))/P(X)

From earlier, we saw that P(X ∩ Y) = P(Y). Thus;

P(Y | X) = P(Y)/P(X)

From the question, P(Y) = 0.9

Since there are 3 relays, then we have;

P(X) = 1 - 0.1³ = 0.999

Thus;

P(Y | X) = 0.9/0.999

P(Y | X) = 0.9009

The  probability that relay 1 functioned is 0.9009

Calculation:

Let us assume X be the event that Current will flow

And,

Let us assume Y be the event that the first relay

So,  Thus, Y ⊂ X.

Now

P(X ∩ Y) = P(Y)

The probability that relay 1 functioned should be

P(Y | X) = (P(Y ∩ X)) ÷ P(X)

From earlier, we saw that P(X ∩ Y) = P(Y).

So,

P(Y | X) = P(Y) ÷ P(X)

Now

P(X) =[tex]1 - 0.1^3[/tex] = 0.999

So,

P(Y | X) = 0.9  ÷0.999

P(Y | X) = 0.9009

learn more about the probability here: https://brainly.com/question/4203858