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The following two-way table shows the distribution of high school students categorized by their grade level and book-type preference.

A 4-column table with 3 rows. Column 1 has entries junior, sophomore, total. Column 2 is labeled fiction with entries 20, 26, 46. Column 3 is labeled nonfiction with entries 22, 13, 35. Column 4 is labeled Total with entries 42, 39, 81. The columns are titled book-type preference and the rows are titled grade level.

Suppose a high school student is selected at random. Let event A = junior and event B = fiction. Are events A and B independent?

Yes, P(A) = P(A|B).
Yes, P(A) = P(B|A).
No, P(A) ≠ P(A|B).
No, P(A) ≠ P(B|A).

[tex]P(A)=\frac{42}{81}\\P(B) =\frac{46}{81}\\P(A B) = \frac{20}{81}\\P(A|B) =\frac{P(A B)}{P(B)} = \frac{20}{46} \neq P(A) => P(AB) = \frac{20}{81} \neq P(A)P(B) = \frac{23}{81}\\\\P(B|A)=\frac{P(AB)}{P(A)}=\frac{20}{42} \neq P(B) => P(AB) = \frac{20}{81} \neq P(A)P(B)= \frac{23}{81}\\\\[/tex]

Note: (AB) means (A ∩ B)

maheshpatelvVT maheshpatelvVT · Answer 2 · 2022-05-14T07:01:12+02:00

The probability of P(A) is not equal to the probability of P(A|B) option third No, P(A) ≠ P(A|B) is correct.

What is probability?

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words the probability is the number that shows the happening of the event.

Two events must be independent of each other, condition must be maintained:

P(A and B) = P(A) × P(B)

The probability:

P(A) = 42/81

P(B) = 46/81

P(AB) = 20/81

P(A|B) = P(AB)/P(B) = 20/46 ≠ P(A) = 20/81 ≠ P(A)P(B) = 23/81

P(B|A) = P(AB)/P(A) = 20/42 ≠ P(B) = 20/81 ≠ P(A)P(B) = 23/81

Thus, the probability of P(A) is not equal to the probability of P(A|B) option third No, P(A) ≠ P(A|B) is correct.

Learn more about the probability here:

brainly.com/question/11234923

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