Question

The record of a weather station shows that out of the past 300 consecutive days, its weather was forecasted correctly 195 times. What is the probability that on a given day selected at random, it was not correct?$$$$

Answer 1

We take selection of a day out of the total 300 days with incorrect forecasting of weather at random as $A$. We find the number of days the weather was foretasted incorrectly as $n\left( A \right)$. We have a total number of days as $n\left( S \right)=300$. The required probability is $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$. $$$$

Complete step by step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring or number of favorable outcomes and $n\left( S \right)$ is the size of the sample space or all possible outcomes, then the probability of the event $A$ occurring is $\dfrac{n\left( A \right)}{n\left( S \right)}$. We are given the question that the record of a weather station shows that out of the past 300 consecutive days, its weather was foretasted correctly 195 times. So the number of times the weather was foretasted incorrectly is $300-195=105$ times.
The event is a selection of a day out of the total 300 days at random. We are asked in the question to find the probability that on that day the weather was foretasted incorrectly. . Let us denote the event of selecting a day with incorrect forecasting as A. So we have a number of days where weather was foretasted incorrectly 105 times which is our number of favorable outcomes. So we have $n\left( A \right)=105$ . We can select the day from 300 days which is all possible outcomes. So we have $n\left( S \right)=300$.
So the required probability is,
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{105}{300}=\dfrac{7}{20}=0.35$

Note: We note that we are asked to find the probability of the selection of a day with incorrect forecasting not the probability of correct or incorrect forecasting in a particular day which is $\dfrac{1}{2}$. We can alternatively solve by first finding the probability of selecting with correct forecasting $B$ as $P\left( B \right)$ and then probability of selecting with incorrect forecasting as $1-P\left( A \right)$ because both the events are mutually exclusive and exhaustive.