What is the probability that the following system works if each unit fails independently with probability p?
____o_____o_____
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_______o________
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____o_____o_____
Note: In the above circuit "o" represents 5 relays in the circuit with propability p
____o_____o_____
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_______o_______|
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____o_____o_____|
Sorry. This is the correct picture
____o_____o_____
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_______o________
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____o_____o_____
This is the correct picture
To calculate the probability that the system works, we need to consider the probability that each relay in the circuit functions properly. Since each unit fails independently with probability p, the probability that a relay functions properly is 1 - p.
Considering that there are 5 relays in the circuit, the probability that a single relay functions properly is (1 - p). Therefore, the probability that all 5 relays in the circuit function properly is (1 - p)^5.
To find the total probability that the system works, we need to assume that the relays are connected in series. In a series circuit, the overall probability is obtained by multiplying the probabilities of each component functioning properly. Therefore, the probability that the system works is (1 - p)^5.
Note that this assumption holds if the system only functions if all the relays work properly, and it fails if any of the relays fail. If the system has additional components or operates in a different way, the calculation may differ.
To determine the probability that the system works, we need to calculate the probability that all the units in the circuit do not fail.
In this circuit, each unit (represented by "o") has a probability p of not failing. Therefore, the probability of a single unit not failing is 1 - p.
Since the units are arranged in a series circuit, all the units must work for the system to work. The units are independent, meaning the failure of one unit does not affect the others.
To calculate the probability that all the units work, we multiply the probabilities of each unit not failing together.
In this case, there are three series pairs of units:
1) The first two units on the left side
2) The second and third units in the middle
3) The last two units on the right side
Let's take the first pair as an example. The probability that both units in this pair do not fail is (1-p) * (1-p).
To calculate the probability that the entire system works, we multiply the probabilities of the three pairs:
(1-p)*(1-p) * (1-p)*(1-p) * (1-p)*(1-p)
Therefore, the probability that the system works is (1-p)^2 * (1-p)^2 * (1-p)^2, which can be simplified as (1-p)^6.