Question

If two events $P(A \cup B) = \dfrac{5}{6}$, $P({A^1}) = \dfrac{5}{6},P(B) = \dfrac{2}{3}$ then A and B are
A). Independent
B). Mutually exclusive
C). Mutually exhaustive
D). Dependent

Answer 1

Hint:
> Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
> If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
> $\dfrac{1}{6}$which means the favorable event is $1$ and the total count is $6$

Complete step-by-step solution:
Since from the given that we have two events A and B, also from the given $P(A \cup B) = \dfrac{5}{6}$, $P({A^1}) = \dfrac{5}{6},P(B) = \dfrac{2}{3}$
Now we know that $P({A^1}) = 1 - P(A)$ (let us assume the overall total probability value is $1$ (this is the most popular concept that used in the probability that the total fraction will not exceed $1$and everything will be calculated under the number $0 - 1$as zero is the least possible outcome and one is the highest outcome)
Hence, we get $P({A^1}) = 1 - P(A) \Rightarrow P(A) = 1 - P({A^1}) = 1 - \dfrac{5}{6} \Rightarrow \dfrac{1}{6}$ thus we have $P(A) = \dfrac{1}{6}$
Now we make use of the probability formula, that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Since we know that $P(A \cup B) = \dfrac{5}{6}$, $P(A) = \dfrac{1}{6},P(B) = \dfrac{2}{3}$ substituting in the above, we get $P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow \dfrac{5}{6} = \dfrac{1}{6} + \dfrac{2}{3} - P(A \cap B)$
Further solving we get the value of the intersection of the two sets, which is $\dfrac{5}{6} = \dfrac{1}{6} + \dfrac{2}{3} - P(A \cap B) \Rightarrow \dfrac{5}{6} = \dfrac{1}{6} + \dfrac{4}{6} - P(A \cap B) \Rightarrow P(A \cap B) = \dfrac{5}{6} - \dfrac{5}{6} = 0$
Thus, we get $P(A \cap B) = 0$
The two probability events are set to be mutually exclusive if their representation of the intersection is zero, which is $P(A \cap B) = 0$
Therefore, the option $2$, The mutually exclusive is correct.

Note: The two probability events are set to be independent events if their representation of the intersection is $P(A \cap B) = P(A) \times P(B)$. But since we got the intersection as zero and hence these options are incorrect.
The general probability formula is $P = \dfrac{F}{T}$where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.