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Answer:
P(E|F)=60% P(E)=60% Events E and F are independent
Step-by-step explanation:
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The events E and F about student being blonde and student being a junior are independent of each other.
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.
Which pair of events are called independent events?
When one event's occurrence or non-occurrence doesn't affect occurrence or non-occurrence of other event, then such events are called independent events.
Symbolically, we have:
Two events A and B are said to be independent if and only if we have:
[tex]P(A \cap B) = P(A)P(B)[/tex]
Comparing it with chain rule will give
[tex]P(A|B) = P(A)\\P(B|A) = P(B)[/tex] , thus, showing that whether one occurred or not, the other one doesn't care about it (independence).
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] and some values of Y are there as [tex]Y_1, Y_2, ... , Y_n[/tex]
List them in title of the rows and left to the columns. There will be [tex]n \times k[/tex] table of 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] (assuming X 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.
The two-way table for this case is:
[tex]\begin{array}{ccccc}\rm &\rm Sophomore&\rm Junior &\rm Senior & \rm Total\\\rm Blonde &10&30&20&60\\\rm Brown&10&20&10&40\\\rm Total & 20& 50 & 30 &100\end{array}[/tex]
Also, it is given that:
- E = event that the student is blonde
- F = event that the student is a junior
From the table, we get:
[tex]n(E \cap F) = 30\\n(E) = 60\\n(F) = 50\\n(S) = 100[/tex]
Thus, we get:
[tex]P(E) = \dfrac{n(E)}{n(S)} = \dfrac{6050}{100} = \dfrac{3}{5}\\\\P(F) = \dfrac{n(F)}{n(S)} = \dfrac{60}{100} = \dfrac{1}{2}\\\\P(E \cap F) = \dfrac{n(E\cap F)}{n(S)} = \dfrac{30}{100} = \dfrac{3}{10}\\\\[/tex]
Thus, we see that:
[tex]P(E \cap F) = \dfrac{3}{10} = \dfrac{3}{5} \times \dfrac{1}{2} = P(E) \times P(F)[/tex]
Thus, E and F are independent events.
Learn more about two-way table here:
https://brainly.com/question/26788374
