Answer :
Answer:
Approximately [tex]0.989[/tex].
Step-by-step explanation:
Based on information provided, construct a table for the joint probability distribution of the two events: rain, and delay:
[tex]\begin{aligned}&& (\text{Rain}) && (\text{No Rain}) \\ (\text{Delay}) && 0.08 && ? && 0.09 \\ (\text{No Delay}) && ? && ? && \\ && 0.1 && &&\end{aligned}[/tex].
Numbers inside the two-by-two square at the center represent the joint probability for each possible pair of events:
- (rain AND delay),
- (no rain AND delay),
- (rain AND no delay), as well as
- (no rain AND no delay).
For example, the number [tex]0.08[/tex] in the table represents (rain AND delay)- the probability that it rains and flight would be delayed.
Numbers on the margin of this table (outside the two-by-two square) represents the probability of an individual event. For example:
- The number [tex]0.09[/tex] in the "delay" row represents the probability that flight would be delayed, regardless of whether it would rain or not.
- The number [tex]0.1[/tex] in the "rain" column represents the probability that it rains, regardless of whether flight would be delayed.
In each row of the table, the sum of numbers in that row of the two-by-two square should be equal to the probability at the margin of that row. The same is true for each column in this table. For example:
- The two probabilities in the first row of the two-by-two square refer to the events (rain AND delay) and (no rain AND delay). The sum of these two probabilities should be equal to the probability at the margin of this row, which represents the probability of (delay).
[tex]P(\text{rain $\land$ delay}) + P(\text{no rain $\land$ delay}) = P(\text{delay})[/tex].
[tex]0.08 + P(\text{no rain $\land$ delay}) = 0.09[/tex].
[tex]P(\text{no rain $\land$ delay}) = 0.01[/tex]. - Similarly, the two probabilities in the first column of the two-by-two square refer to the events (rain AND delay) and (rain AND no delay). The sum of these two probabilities should be equal to the probability at the margin of this column, which represents the probability of (rain).
[tex]P(\text{rain $\land$ delay}) + P(\text{rain $\land$ no delay}) = P(\text{rain})[/tex].
[tex]0.08 + P(\text{rain $\land$ no delay}) = 0.1[/tex].
[tex]P(\text{rain $\land$ no delay}) = 0.02[/tex].
Fill these values into the table:
[tex]\begin{aligned}&& (\text{Rain}) && (\text{No Rain}) \\ (\text{Delay}) && 0.08 && {\bf 0.01} && 0.09 \\ (\text{No Delay}) && {\bf 0.02} && ? && \\ && 0.1 && &&\end{aligned}[/tex].
Entries in the two-by-two square at the center of this table represent the likelihood of all possible values for the joint probability distribution between the two events: rain and delay. Like other probability distributions, the probabilities of all these possible values should add up to [tex]1[/tex]:
[tex]\begin{aligned}& P(\text{rain $\land$ delay}) \\ +& P(\text{no rain $\land$ delay}) \\ +& P(\text{rain $\land$ no delay}) \\ +& P(\text{no rain $\land$ no delay}) \\ =\; & 1 \end{aligned}[/tex].
[tex]0.08 + 0.01 + 0.02 + P(\text{no rain $\land$ no delay}) = 1[/tex].
Hence:
[tex]\begin{aligned} & P(\text{no rain $\land$ no delay}) \\ =\; & 1- 0.08 - 0.01 - 0.02 \\ =\; & 0.89\end{aligned}[/tex].
In other words, the joint probability that it would not rain and the flight would not be delayed would be [tex]0.89[/tex]. To find [tex]P(\text{no rain $|$ no delay})[/tex], the conditional probability of no delay "given" that it would not rain, divide the joint probability [tex]P(\text{no rain $\land$ no delay})[/tex] by the probability of the premise that it would not rain [tex]P(\text{no rain})[/tex]:
[tex]\begin{aligned} & P(\text{no rain $|$ no delay}) \\ =\; & \frac{P(\text{no rain $\land$ no delay})}{P(\text{no rain})} \\ =\; & \frac{P(\text{no delay $\land$ no rain})}{P(\text{no rain})} \\ =\; & \frac{P(\text{no delay $\land$ no rain})}{1 - P(\text{rain})} \\ =\; & \frac{0.89}{1 - 0.1} \\ =\; & \frac{89}{90} \\ \approx\; & 0.989\end{aligned}[/tex].
Note that given [tex]P(\text{rain}) = 0.1[/tex], the probability of the complementary event of no rain would be [tex]P(\text{no rain}) = 1- P(\text{rain})[/tex].
In other words, given that it would not be raining, the probability that the flight would not be delayed would be approximately [tex]0.989[/tex].
