If the fire alarm rings, What is the conditional probability that there is a fire?

Question

Here is the statement:- On 1 day in 1000, there is a fire and the fire alarm rings.

On 1 day in 100, there is no fire and the fire alarm rings (false alarm)

On 1 day in 10,000, there is a fire and the fire alarm does not ring (defective alarm)

On 9,889 days out of 10,000, there is no fire and the fire alarm does not ring.

Written p(there is a fire | fire alarm rings)

I am stuck at figuring out the probability I have applied conditional probability and get p(a) =0.0001 assuming p(b) be true but it is not right way way to solve can someone help me to solve this question?

Answer 1

Make a frequency table. Let $F$ represent the event of a fire, and $\bar F$ represent no fire. Let $A$ represent the event that the alarm rings, and $\bar A$ represent no alarm. Then convert all rates into frequencies over the common denominator $10000$, so for example, $1$ in $100$ is the same as $100$ in $10000$.

$$\begin{array}{c|c|c|c} & A & \bar A & \\ \hline F & 10 & 1 & 11 \\ \hline \bar F & 100 & 9889 & 9989 \\ \hline & 110 & 9890 & 10000 \\ \end{array}$$

The marginal totals are just the sums of the rows and columns. Now can you use this table to compute $\Pr[F \mid A]$?

Answer 2

Best you write down, what you have. Your events are "there is a fire today" ($F$) and "it rings today" ($R$) and their complements. You are given the probabilities $\mathbb P (F \cap R)$, $\mathbb P (F \cap R^c)$ and $\mathbb P (F^c\cap R)$. You want to compute $\mathbb P (F | R)$. By definition, this is \begin{align*} \mathbb P (F|R) = \frac{\mathbb P (F\cap R)}{\mathbb P (R)} \end{align*}
So you only need to find out $\mathbb P (R)$. Here you can use, that probability measures are additive on disjoint events. \begin{align*} \mathbb P (R) = \mathbb P (F\cap R) + \mathbb P (F^c \cap R)\end{align*}

Answer 3

So the conditional probability of a fire given that the fire alarm rings is 0.084 or 8.4%. Therefore, the answer is 8.4%