What is the maximum probability of failure on demand for a smart transmitter with a failure rate of 0.05 failures/year tested twice per year?

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Multiple Choice

What is the maximum probability of failure on demand for a smart transmitter with a failure rate of 0.05 failures/year tested twice per year?

To determine the maximum probability of failure on demand for a smart transmitter tested twice per year with a failure rate of 0.05 failures/year, we need to analyze how the testing frequency and failure rate interact. The failure rate indicates that, on average, the device will fail 0.05 times per year. Since the transmitter is tested twice a year, we can calculate the probability of failure on demand by considering the number of opportunities for testing within the year. Firstly, if the transmitter fails at a rate of 0.05 per year, the probability of it not failing in a year is calculated by taking the exponent of the negative failure rate: \[ P(\text{not failing in one year}) = e^{-\lambda} \] where \(\lambda\) is the failure rate (0.05 in this case). This gives us: \[ P(\text{not failing in one year}) = e^{-0.05} \] Calculating this results in approximately 0.951. Next, we can find the probability of failing at least once during the year. This is the complement of the probability of not failing at all: \[ P(\text{at least one failure in one year}) = 1 -

To determine the maximum probability of failure on demand for a smart transmitter tested twice per year with a failure rate of 0.05 failures/year, we need to analyze how the testing frequency and failure rate interact.

The failure rate indicates that, on average, the device will fail 0.05 times per year. Since the transmitter is tested twice a year, we can calculate the probability of failure on demand by considering the number of opportunities for testing within the year.

Firstly, if the transmitter fails at a rate of 0.05 per year, the probability of it not failing in a year is calculated by taking the exponent of the negative failure rate:

[ P(\text{not failing in one year}) = e^{-\lambda} ]

where (\lambda) is the failure rate (0.05 in this case). This gives us:

[ P(\text{not failing in one year}) = e^{-0.05} ]

Calculating this results in approximately 0.951.

Next, we can find the probability of failing at least once during the year. This is the complement of the probability of not failing at all:

[ P(\text{at least one failure in one year}) = 1 -

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