Consider the following system considering of two switches I and II between two points A and B:

A signal is sent from the point A to point B and is received at B if both the switches I and II are closed. It is assumed that the probabilities of I and II being closed are 0.8 and 0.6 respectively and that P [II is closed] I is closed] = P [II is closed].

Find:

i) The probability that signal is received at B.

ii) The conditional probability that switch I was open, given that the signal was not received at B,

iii) The conditional probability that switch II was open, given that the signal was not received at B.

Let's go step by step.

Given:

• Probability of switch I being closed: P(I closed)=0.8P(I \text{ closed}) = 0.8
• Probability of switch II being closed: P(II closed)=0.6P(II \text{ closed}) = 0.6
• The events for switches I and II being closed are independent, i.e., P(II closed∣I closed)=P(II closed)P(II \text{ closed} \mid I \text{ closed}) = P(II \text{ closed}).

i) Probability that the signal is received at B:

The signal will be received at B only if both switches I and II are closed. Since the switches are independent, the probability of both being closed is the product of their individual probabilities:

P(signal received at B)=P(I closed)×P(II closed)=0.8×0.6=0.48P(\text{signal received at B}) = P(I \text{ closed}) \times P(II \text{ closed}) = 0.8 \times 0.6 = 0.48

ii) Conditional probability that switch I was open, given that the signal was not received at B:

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