Question

use the values in the cross-tabulation table to solve the equations given.

\td\te\tf
a\t4\t6\t11
b\t11\t6\t5
c\t7\t2\t5

(round your answers to 4 decimal places, e.g. 0.2156.)

a. ( p(a cup d) = )

b. ( p(e cup b) = )

c. ( p(d cup e) = )

d. ( p(c cup f) = )
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Answer:

First, we need to find the total number of elements in the table. Let's sum all the values:

For row A: \( 4 + 6 + 11 = 21 \)

For row B: \( 11 + 6 + 5 = 22 \)

For row C: \( 7 + 2 + 5 = 14 \)

Total \( N = 21 + 22 + 14 = 57 \)

Part a: \( P(A \cup D) \)

We use the formula for the probability of the union of two events: \( P(A \cup D) = P(A) + P(D) - P(A \cap D) \)

• \( P(A) \): Number of elements in A divided by total. Elements in A: \( 4 + 6 + 11 = 21 \), so \( P(A) = \frac{21}{57} \)
• \( P(D) \): Number of elements in D divided by total. Elements in D: \( 4 + 11 + 7 = 22 \), so \( P(D) = \frac{22}{57} \)
• \( P(A \cap D) \): Number of elements in both A and D (which is the cell A-D) divided by total. That's \( 4 \), so \( P(A \cap D) = \frac{4}{57} \)

Now calculate:

\( P(A \cup D) = \frac{21}{57} + \frac{22}{57} - \frac{4}{57} = \frac{21 + 22 - 4}{57} = \frac{39}{57} \approx 0.6842 \)

Part b: \( P(E \cup B) \)

Using the union formula: \( P(E \cup B) = P(E) + P(B) - P(E \cap B) \)

• \( P(B) \): Elements in B: \( 11 + 6 + 5 = 22 \), so \( P(B) = \frac{22}{57} \)
• \( P(E) \): Elements in E: \( 6 + 6 + 2 = 14 \), so \( P(E) = \frac{14}{57} \)
• \( P(E \cap B) \): Elements in both E and B (cell B-E) is \( 6 \), so \( P(E \cap B) = \frac{6}{57} \)

Calculate:

\( P(E \cup B) = \frac{22}{57} + \frac{14}{57} - \frac{6}{57} = \frac{22 + 14 - 6}{57} = \frac{30}{57} \approx 0.5263 \)

Part c: \( P(D \cup E) \)

Union formula: \( P(D \cup E) = P(D) + P(E) - P(D \cap E) \)

• \( P(D) = \frac{22}{57} \) (from part a)
• \( P(E) = \frac{14}{57} \) (from part b)
• \( P(D \cap E) \): Elements in both D and E (cells A-D, B-D, C-D? No, wait, D and E are columns. So D ∩ E is the cells where row is any and column is D and E? Wait no, actually, D and E are columns, so the intersection of D and E would be the cells that are in both D and E, but since D and E are distinct columns, the only way is the cells that are in D and E, but actually, for events D (column D) and E (column E), the intersection is the cells that are in both column D and column E, but since a cell is in one column, the intersection of D and E is empty? Wait no, no. Wait, D is the event "being in column D", E is "being in column E". So a cell can't be in both column D and column E, so \( P(D \cap E) = 0 \)? Wait no, that's not right. Wait, no, D and E are columns, so the elements in D are the cells in column D: (A,D)=4, (B,D)=11, (C,D)=7. Elements in E are (A,E)=6, (B,E)=6, (C,E)=2. So the intersection of D and E is the set of cells that are in both D and E, but since a cell is in one column, there are no cells in both D and E. So \( P(D \cap E) = 0 \). Wait, but let's check the formula again. Wait, no, actually, D and E are mutually exclusive? No, wait, no. Wait, D is the event of being in column D, E is column E. So a single cell can't be in both columns, so they are mutually exclusive. So \( P(D \cup E) = P(D) + P(E) \)

Wait, let's recalculate:

Elements in D: \( 4 + 11 + 7 = 22 \)

Elements in E: \( 6 + 6 + 2 = 14 \)

Since D and E are columns, no overlap (a cell is in one column), so \( P(D \cap E) = 0 \)

Thus, \( P(D \cup E) = \frac{22}{57} + \frac{14}{57} = \frac{36}{57} \approx 0.6316 \)

Part d: \( P(C \cup F) \)

Union formula: \( P(C \cup F) = P(C) + P(F) - P(C \cap F) \)

• \( P(C) \): Elements in C: \( 7 + 2 + 5 = 14 \), so \( P(C) = \frac{14}{57} \)
• \( P(F) \): Elements in F: \( 11 + 5 + 5 = 21 \), so \( P(F) = \frac{21}{57} \)
• \( P(C \cap F) \): Elements in both C and F (cell C-F) is \( 5 \), so \( P(C \cap F) = \frac{5}{57} \)

Calculate:

\( P(C \cup F) = \frac{14}{57} + \frac{21}{57} - \frac{5}{57} = \frac{14 + 21 - 5}{57} = \frac{30}{57} \approx 0.5263 \)

Final Answers:

a. \( \boldsymbol{0.6842} \)

b. \( \boldsymbol{0.5263} \)

c. \( \boldsymbol{0.6316} \)

d. \( \boldsymbol{0.5263} \)