Question

The function, \( r(t) \), signifies the likelihood that a component has survived up to time \( t \) and fails in the next instant of time.
\[ r(t) = f(t) / (1 - F(t)) \]
where \( f(t) \) denotes the probability density function for time to failure with corresponding cumulative distribution function given by,
\[ F(t) = 1 - e^{-at^b}, \quad a > 0, b > 0 \]
Which ONE of the following represents \( r(t) \)?

• A. \( a \times b \times t^{b-1} \times e^{-at^b} \)
• B. \( a \times b \times t^{b-1} \)
• C. \( 1 - (a \times b \times t^{b-1}) \)
• D. \( 1 - (a \times b \times t^{b-1} \times e^{-at^b}) \)

Hint:

Differentiate \( F(t) \) to get \( f(t) \), then divide by \( 1-F(t) \); the exponential terms cancel.

Answer 1

The Correct Option is B

Step 1: Identify what is being asked:
We are given the cumulative distribution function (CDF) of the time to failure, \( F(t) \), and we must find the hazard rate function \( r(t) \), which is defined as the ratio of the density function \( f(t) \) to the reliability function \( 1 - F(t) \).
This kind of function is called the hazard rate or failure rate in reliability engineering.

Step 2: Differentiate the CDF to get the PDF:
The probability density function is the time derivative of the CDF.
Since \( F(t) = 1 - e^{-at^b} \), we differentiate term by term using the chain rule.
\[ f(t) = \frac{d}{dt}\left(1 - e^{-at^b}\right) = -e^{-at^b} \cdot \frac{d}{dt}(-at^b) = a b t^{b-1} e^{-at^b} \]

Step 3: Compute the reliability function and form the ratio:
The reliability function is simply \( 1 - F(t) \).
\[ 1 - F(t) = 1 - \left(1 - e^{-at^b}\right) = e^{-at^b} \]
Now substitute \( f(t) \) and \( 1 - F(t) \) into the hazard rate expression.
\[ r(t) = \frac{f(t)}{1 - F(t)} = \frac{a b t^{b-1} e^{-at^b}}{e^{-at^b}} \]
The exponential terms in the numerator and denominator are identical, so they cancel completely, leaving a simple power law in \( t \).

Step 4: Check the remaining options:
Option (A) still carries the exponential term \( e^{-at^b} \), which is actually \( f(t) \) itself, not the ratio \( r(t) \), so it is wrong.
Options (C) and (D) subtract the hazard-like term from 1, which has no basis in the derivation above and does not follow from dividing \( f(t) \) by \( 1 - F(t) \), so both are incorrect.

Final Answer:
Cancelling the exponential terms gives a clean result independent of the survival probability. \[ \boxed{r(t) = a \times b \times t^{b-1}} \]