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.
Show 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.
Step 1: Understanding the Concept:
This is a conditional probability problem. We need to find the probability of a specific failure mode given that the overall instrument has failed. Step 2: Detailed Explanation:
Let \(A\) be the event that the 1st unit functions and \(B\) be the event that the 2nd unit functions.
\(P(A) = 0.9 \implies P(A') = 0.1\).
\(P(B) = 0.8 \implies P(B') = 0.2\).
The instrument operates only if both function: \(P(\text{Operates}) = P(A \cap B) = 0.9 \times 0.8 = 0.72\).
The instrument fails if it does not operate: \(P(F) = 1 - 0.72 = 0.28\).
We want the probability \(p = P(A' \cap B | F)\).
By definition of conditional probability:
\[ p = \frac{P(A' \cap B \cap F)}{P(F)} \]
Since \(A' \cap B\) implies the instrument has failed, \(A' \cap B \cap F = A' \cap B\).
\[ P(A' \cap B) = P(A')P(B) = 0.1 \times 0.8 = 0.08 \]
\[ p = \frac{0.08}{0.28} = \frac{8}{28} = \frac{2}{7} \]
The value required is \(98p = 98 \times \frac{2}{7} = 14 \times 2 = 28\). Step 3: Final Answer:
The value is 28.
