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

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.