Math Problem Statement

For each set of probabilities, determine whether the events and are independent or dependent.

(If necessary, consult a list of formulas.)

Probabilities Independent Dependent

(a) P(A) = 1/6; P(B) = 1/5; P(B|A) = 1/5

(b) P(A) = 1/5; P(B) = 1/2; P(A|B) = 1/10

(c) P(A) = 1/5; P(B) = 1/3; P(A and B) = 1/2

(d) P(A) = 1/4; P(B) = 1/9; P(A and B) = 1/5

Solution

To determine whether events AA and BB are independent or dependent, we use the definition of independence:

P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B)

If this equality holds, then AA and BB are independent; otherwise, they are dependent.


(a) P(A)=16,P(B)=15,P(BA)=15P(A) = \frac{1}{6}, P(B) = \frac{1}{5}, P(B|A) = \frac{1}{5}

Using the conditional probability formula:

P(BA)=P(AB)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}

15=P(AB)1/6\frac{1}{5} = \frac{P(A \cap B)}{1/6}

Solving for P(AB)P(A \cap B):

P(AB)=15×16=130P(A \cap B) = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30}

Now, checking independence:

P(A)P(B)=16×15=130P(A) \cdot P(B) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}

Since P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B), AA and BB are independent.


(b) P(A)=15,P(B)=12,P(AB)=110P(A) = \frac{1}{5}, P(B) = \frac{1}{2}, P(A|B) = \frac{1}{10}

Using:

P(AB)=P(AB)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

110=P(AB)1/2\frac{1}{10} = \frac{P(A \cap B)}{1/2}

Solving for P(AB)P(A \cap B):

P(AB)=110×12=120P(A \cap B) = \frac{1}{10} \times \frac{1}{2} = \frac{1}{20}

Checking independence:

P(A)P(B)=15×12=110P(A) \cdot P(B) = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10}

Since P(AB)P(A)P(B)P(A \cap B) \neq P(A) \cdot P(B) (because 120110\frac{1}{20} \neq \frac{1}{10}), AA and BB are dependent.


(c) P(A)=15,P(B)=13,P(AB)=12P(A) = \frac{1}{5}, P(B) = \frac{1}{3}, P(A \cap B) = \frac{1}{2}

Checking independence:

P(A)P(B)=15×13=115P(A) \cdot P(B) = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}

Since P(AB)P(A)P(B)P(A \cap B) \neq P(A) \cdot P(B) (because 12115\frac{1}{2} \neq \frac{1}{15}), AA and BB are dependent.


(d) P(A)=14,P(B)=19,P(AB)=15P(A) = \frac{1}{4}, P(B) = \frac{1}{9}, P(A \cap B) = \frac{1}{5}

Checking independence:

P(A)P(B)=14×19=136P(A) \cdot P(B) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36}

Since P(AB)P(A)P(B)P(A \cap B) \neq P(A) \cdot P(B) (because 15136\frac{1}{5} \neq \frac{1}{36}), AA and BB are dependent.


Final Answers:

(a) Independent
(b) Dependent
(c) Dependent
(d) Dependent


Would you like a step-by-step breakdown for any of these?

Related Questions:

  1. How do you calculate conditional probability using P(AB)P(A|B)?
  2. What is the difference between independent and mutually exclusive events?
  3. How does Bayes' theorem relate to conditional probability?
  4. Can two events be both independent and mutually exclusive?
  5. What real-world examples involve dependent probability?

Tip: When determining independence, always compare P(AB)P(A \cap B) with P(A)P(B)P(A) \cdot P(B). If they are equal, the events are independent; otherwise, they are dependent.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Independence of Events

Formulas

P(A and B) = P(A) * P(B) (for independent events)
P(B|A) = P(A and B) / P(A) (conditional probability formula)
P(A|B) = P(A and B) / P(B) (conditional probability formula)

Theorems

Definition of Independent Events

Suitable Grade Level

Grades 10-12