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Sagot :
The two events out of the listed events which are independent events are given by: Option A: A and C
What is chain rule in probability?
For two events A and B, by chain rule, we have:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)[/tex]
What is law of total probability?
Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as
[tex]B_i \: ; i \in \{1,2,3.., n\}[/tex]
Then, suppose there is event A in sample space.
Then probability of A's occurrence can be given as
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i)[/tex]
Using the chain rule, we get
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)[/tex]
How to form two-way table?
Suppose two dimensions are there, viz X and Y. Some values of X are there as [tex]X_1, X_2, ... , X_n[/tex] values of Y are there as [tex]Y_1, Y_2, ..., Y_k[/tex]rows and left to the columns. There will be [tex]n \times k[/tex]values will be formed(excluding titles and totals), such that:
Value([tex]i^{th}[/tex] row, [tex]j^{th}[/tex] column) = Frequency for intersection of [tex]X_i[/tex] and [tex]Y_j[/tex]values are going in rows, and Y values are listed in columns).
Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.
For n = 2, and k = 2, the table would look like:
[tex]\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}[/tex]
where S denotes total of totals, also called total frequency.
n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
How to find if two events are independent?
Suppose that two events are denoted by A and B.
They are said to be independent event if and only if:
[tex]P(A \cap B) = P(A)P(B)[/tex]
The given frequency table is:
[tex]\begin{array}{ccccc} &\text{Public}&\text{Own}&\text{Others}&\text{Total}\\\text{Male}&12&20&4&36\\\text{Female}&8&10&6&24\\\text{Total}&20&30&10&60\end{array}[/tex]
The probability table for the same labels would be:
[tex]\begin{array}{ccccc} &\text{Public}&\text{Own}&\text{Others}&\text{Total}\\\text{Male}&12/60&20/60&4/60&36/60\\\text{Female}&8/60&10/60&6/60&24/60\\\text{Total}&20/60&30/60&10/60&1\end{array}[/tex]
The events A, B,C,D and E are given as:
- A: The employee is male.
- B: The employee is female.
- C: The employee takes public transportation.
- D: The employee takes his/her own transportation.
- E: The employee takes some other method of transportation.
Checking all the listed options one by one, for them being independent:
- Case 1: A and C
P(A) = P(The employee is male. ) = 36/60
P(C) = P(The employee takes public transportation.) = 20/60[tex]P(A \cap C) = 12/60 \\\\ P(A)P(C) = \dfrac{36 \times 20}{60^2} = 12/60[/tex]
[tex]P(A \cap C) = P(A)P(C)[/tex]
Thus, A and C are independent events.
- Case 2: A and D
P(A) = P(The employee is male. ) = 36/60
P(D) = P(The employee takes his/her own transportation.) = 30/60
[tex]P(A\cap D) = 20/60\\\\P(A)P(D) = \dfrac{30 \times 36}{60^2} = 12/60 \neq P(A \cap D)[/tex]
Thus, A and D are not independent events.
- Case 3: B and D
P(B) = P(The employee is female. ) = 24/60
P(D) = P(The employee takes his/her own transportation.) = 30/60
[tex]P(B \cap D) = 10/60 \neq P(B)P(D)=\dfrac{24 \times 30}{60^2} = 12/60[/tex]
Thus, B and D are not independent events.
- Case 4: B and E
P(B) = P(The employee is female. ) = 24/60
P(E) = P(The employee takes some other method of transportation.) = 10/60
[tex]P(B \cap E) = 6/60 \neq P(B)P(E)= \dfrac{24 \times 10}{60^2} = 4/60[/tex]
Thus, B and E are not independent events.
Thus, the two events out of the listed events which are independent events are given by: Option A: A and C
Learn more about independent events here:
https://brainly.com/question/3898488
Answer:
First option!!
"A and C"
Step-by-step explanation:
Edge 2022
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