Question

Problem 2. A fuse (or circuit breaker) can trip under normal current conditions with probability 0.01. It trips under overloaded conditions with probability 0.9999. We assume that the current is normal 98% of the time. (a) What is the probability that the fuse will trip? (b) If the fuse has tripped, find the conditional probability that the current was normal.

Answer 1

Solution:

Let's consider the experiment 1000 times. We are given that the current is normal 98% of the time.

Therefore, the number of times, we will see a normal current $\dpi{100} \small =1000 \times 0.98=980$

And the number of times, we will see a non-normal current $\dpi{100} \small =1000-980=20$

We are also given that the fuse can trip under normal conditions with probability 0.01.

Therefore, the number of times the fuse can trip under normal conditions $\dpi{100} \small =980 \times 0.01=9.8$

And the number of times the fuse can not trip under normal conditions $\dpi{100} \small =980-9.8=970.2$

We are also given that the fuse can trip under overload conditions with probability 0.9999

Therefore, the number of times a fuse can trip under overload conditions $\dpi{100} \small =20 \times 0.9999=19.998$

And the number of times the fuse can not trip under overload conditions $\dpi{100} \small =20-19.998=0.002$

Therefore, we can express this information in below table as:

Trip No Trip Total 9.8970.2980 Normal Overloaded19.9980.002 Total 20 29.798 970.2021000

(a) What is the probability that the fuse will trip?

Answer: The probability that the fuse will trip is:

$\dpi{100} \small \boldsymbol{\frac{29.798}{1000}=0.0298 }$

(b) If the fuse has tripped, find the conditional probability that the current was normal?

Answer: The conditional probability that the current was normal given the fuse had tripped is:

$\dpi{100} \small \boldsymbol{\frac{9.8}{29.798}=0.3289}$