Question

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is \(0.9\) and that of the second unit is \(0.8\). The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is \(p\), then \(98p\) is equal to \dots\dots.

Hint:

For conditional probability \(P(X|F)\), always ensure that event \(X\) is actually a subset of the condition \(F\). Here, failing one unit is a specific way the whole instrument fails.

Answer 1

Correct Answer: 28

The problem revolves around conditional probability. To find \(p\), the probability that only the first unit failed while the second one functioned, given the entire instrument failed to operate, we first determine the probabilities of different scenarios for failures.
The instrument fails if:
1. Both units fail.
2. The first unit fails and the second unit operates.
3. The first unit operates and the second unit fails.
Calculate probabilities:
- Probability that both units fail: \( (1-0.9) \times (1-0.8) = 0.1 \times 0.2 = 0.02 \)
- Probability that first unit fails and second unit operates: \( (1-0.9) \times 0.8 = 0.1 \times 0.8 = 0.08 \)
- Probability that first unit operates and second unit fails: \( 0.9 \times (1-0.8) = 0.9 \times 0.2 = 0.18 \)
Total probability of instrument failure: \( 0.02 + 0.08 + 0.18 = 0.28 \)
Conditional Probability:
\( p = \frac{\text{Probability that the first unit fails and second operates}}{\text{Total probability of failure}}=\frac{0.08}{0.28}=\frac{2}{7} \)
Finally, we find \( 98p = 98 \times \frac{2}{7} = 28 \).
This computed value \( 28 \) falls within the specified range of \(28,28\). Thus, our solution is verified.