An electronic device is made up of two components A and B such that the device would work satisfactorily as long as at least one of the components works. The probability of failure of component A is 0.02 and that of B is 0.1 in some fixed period of time. If the components work independently, find the probability that the device will work satisfactorily during the period.

#### Solution

Let $A$ be the event that component A work. Let $B$ be the event that component B work.

The probability of failure of component A is 0.02, i.e., `$P(A^\prime) = 0.02$`

and the probability of failure of component B is 0.1, i.e., `$P(B^\prime) = 0.1$`

.

Then `$P(A)= 1-P(A^\prime) = 0.98$`

and, $P(B) = 1- `P(B^\prime)=0.9$`

.

The probability that the device will work satisfactorily during the period is given by `$P(A\cap B)$`

.

As process of occurrence of both the event is independent to each other, we use

`$P(A\cap B) = P(A)*P(B)$`

The probability that the device will work satisfactorily during the period is `$P(A\cap B)$`

.

That is, `$P(A\cap B) = 0.98*0.9 = 0.882$`

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#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators