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

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.