Question

A and B are two independent events such that $P(A \cup {B^1}) = 0.8$ and $P(A) = 0.3$ , then $P(B)$ is
$A)\dfrac{2}{7}$
$B)\dfrac{2}{3}$
$C)\dfrac{3}{8}$
$D)\dfrac{1}{8}$

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 $6$
> In this problem, they represent the probability in the division which means $0.8 = \dfrac{8}{{10}}$ or with any division value of $0.8$ , remember all values are the same.
Formula used: To find the union of the two sets, we generally use the probability of $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and similarly in this problem we are going to use $P(A \cup {B^1}) = P(A) + P({B^1}) - P(A \cap {B^1})$

Complete step-by-step solution:
Since from the given that we have, A and B are two independent events such that $P(A \cup {B^1}) = 0.8$ and $P(A) = 0.3$ with this information, we have to find $P(B)$
The two probability events are set to be independent events if their representation of the intersection is $P(A \cap B) = P(A) \times P(B)$.
Hence, we will rewrite the given formula as $P(A \cup {B^1}) = P(A) + P({B^1}) - P(A \cap {B^1}) \Rightarrow P(A \cup {B^1}) = P(A) + P({B^1}) - P(A) \times P({B^1})$
Given that $P(A \cup {B^1}) = 0.8$ and $P(A) = 0.3$.
Substituting these values, we get $P(A \cup {B^1}) = P(A) + P({B^1}) - P(A) \times P({B^1}) \Rightarrow 0.8 = 0.3 + P({B^1}) - 0.3 \times P({B^1})$ $ \Rightarrow 0.8 - 0.3 = P({B^1}) - 0.3.P({B^1})$
Taking the common values, we get $ \Rightarrow 0.5 = P({B^1})[1 - 0.3]$
Further solving we get $ \Rightarrow 0.5 = P({B^1})[1 - 0.3] \Rightarrow 0.5 = P({B^1})[0.7]$
Since $0.5 = \dfrac{5}{{10}},0.7 = \dfrac{7}{{10}}$ then we get $0.5 = P({B^1})[0.7] \Rightarrow \dfrac{5}{{10}} = P({B^1})[\dfrac{7}{{10}}] \Rightarrow P({B^1}) = \dfrac{5}{7}$
Now we know that $P(B) = 1 - P({B^1})$ (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)
Thus, we have, $P({B^1}) = \dfrac{5}{7} \Rightarrow P(B) = 1 - P({B^1}) \Rightarrow 1 - \dfrac{5}{7} = \dfrac{2}{7}$
Therefore, the option $A)\dfrac{2}{7}$ is correct.

Note: The two probability events are set to be mutually exclusive if their representation of the intersection is zero, which is $P(A \cap B) = 0$
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.