Let A and B be two events such that $P(\overline{A \cup B}) = \dfrac{1}{6}$ , P(A ∩ B) = $\dfrac{1}{4}$ and P(A̅) = $\dfrac{1}{4}$ , where A̅ stands for complement of event A. Then, events A and B are:

Question

Let A and B be two events such that $P(\overline{A \cup B}) = \dfrac{1}{6}$ , P(A &cap; B) =  $\dfrac{1}{4}$ and P(A̅) =  $\dfrac{1}{4}$ , where A̅ stands for complement of event A. Then, events A and B are:

Prepp Team · Accepted Answer

Analyzing Probability of Events A and B

The problem provides us with certain probabilities related to two events, A and B, and asks us to determine if these events are equally likely, mutually exclusive, or independent based on the given information.

We are given the following probabilities:

Probability of the complement of the union of A and B: $P(\overline{A \cup B}) = \dfrac{1}{6}$
Probability of the intersection of A and B: $P(A \cap B) = \dfrac{1}{4}$
Probability of the complement of event A: $P(\overline{A}) = \dfrac{1}{4}$

We need to use these values to find $P(A)$ and $P(B)$ and then check the conditions for equally likely, mutually exclusive, and independent events.

Calculating Probabilities P(A) and P(B)

First, let's find the probability of event A, $P(A)$. We know that $P(A) = 1 - P(\overline{A})$.

$P(A) = 1 - P(\overline{A}) = 1 - \dfrac{1}{4} = \dfrac{4 - 1}{4} = \dfrac{3}{4}$

So, $P(A) = \dfrac{3}{4}$.

Next, let's find the probability of the union of A and B, $P(A \cup B)$. We know that $P(\overline{A \cup B}) = 1 - P(A \cup B)$.

$P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - \dfrac{1}{6} = \dfrac{6 - 1}{6} = \dfrac{5}{6}$

So, $P(A \cup B) = \dfrac{5}{6}$.

Now we can find the probability of event B, $P(B)$, using the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the values we know:

$\dfrac{5}{6} = \dfrac{3}{4} + P(B) - \dfrac{1}{4}$

Combine the terms on the right side:

$\dfrac{5}{6} = \dfrac{3}{4} - \dfrac{1}{4} + P(B)$

$\dfrac{5}{6} = \dfrac{2}{4} + P(B)$

$\dfrac{5}{6} = \dfrac{1}{2} + P(B)$

Solve for $P(B)$:

$P(B) = \dfrac{5}{6} - \dfrac{1}{2}$

To subtract, find a common denominator, which is 6:

$P(B) = \dfrac{5}{6} - \dfrac{1 \times 3}{2 \times 3} = \dfrac{5}{6} - \dfrac{3}{6} = \dfrac{5 - 3}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

So, $P(B) = \dfrac{1}{3}$.

Checking Properties of Events A and B

Now that we have $P(A)$ and $P(B)$, we can check if the events A and B are equally likely, mutually exclusive, or independent.

Equally Likely Events

Events A and B are equally likely if $P(A) = P(B)$.

We found $P(A) = \dfrac{3}{4}$ and $P(B) = \dfrac{1}{3}$.

Since $\dfrac{3}{4} \neq \dfrac{1}{3}$, events A and B are not equally likely.

Mutually Exclusive Events

Events A and B are mutually exclusive if they cannot happen at the same time, which means their intersection is empty. In terms of probability, this means $P(A \cap B) = 0$.

We are given $P(A \cap B) = \dfrac{1}{4}$.

Since $P(A \cap B) = \dfrac{1}{4} \neq 0$, events A and B are not mutually exclusive.

Independent Events

Events A and B are independent if the occurrence of one does not affect the probability of the other. In terms of probability, this means $P(A \cap B) = P(A) \times P(B)$.

We have $P(A \cap B) = \dfrac{1}{4}$.

Let's calculate $P(A) \times P(B)$:

$P(A) \times P(B) = \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{3 \times 1}{4 \times 3} = \dfrac{3}{12} = \dfrac{1}{4}$

Since $P(A \cap B) = \dfrac{1}{4}$ and $P(A) \times P(B) = \dfrac{1}{4}$, we see that $P(A \cap B) = P(A) \times P(B)$.

Therefore, events A and B are independent.

Conclusion

Based on our calculations:

Events A and B are not equally likely ($P(A) \neq P(B)$).
Events A and B are not mutually exclusive ($P(A \cap B) \neq 0$).
Events A and B are independent ($P(A \cap B) = P(A) \times P(B)$).

Thus, events A and B are independent but not equally likely.

Revision Table: Probability Concepts

Concept	Definition	Condition
Equally Likely Events	Events with the same probability of occurrence.	$P(A) = P(B)$
Mutually Exclusive Events	Events that cannot occur at the same time (no common outcomes).	$A \cap B = \emptyset$ or $P(A \cap B) = 0$
Independent Events	Events where the outcome of one does not affect the outcome of the other.	$P(A \cap B) = P(A) \times P(B)$ or $P(A\|B) = P(A)$ if $P(B) > 0$ or $P(B\|A) = P(B)$ if $P(A) > 0$

Additional Information: Probability Rules

Here are some fundamental probability rules used in solving this problem:

Complement Rule: The probability of an event not happening is 1 minus the probability of it happening. $P(\overline{A}) = 1 - P(A)$
Union Rule: The probability of either event A or event B (or both) occurring is the sum of their individual probabilities minus the probability of their intersection. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
De Morgan's Laws (for probability): The probability of the complement of the union is 1 minus the probability of the union. $P(\overline{A \cup B}) = 1 - P(A \cup B)$. Also, the probability of the complement of the intersection is 1 minus the probability of the intersection. $P(\overline{A \cap B}) = 1 - P(A \cap B)$. These are helpful when dealing with complements of combined events.

Understanding these basic rules is crucial for solving probability problems involving multiple events.

Prepp Team · Answer

Equally likely but not independent

Prepp Team · Answer

Mutually exclusive and independent

Prepp Team · Answer

Equally likely and mutually exclusive