Question

If A and B are two events such that $P\left( A\cup B \right)+P\left( A\cap B \right)=\dfrac{7}{8}$ and $P\left( A \right)=2P\left( B \right)$, then $P\left( A \right)=$ ?
(a) $\dfrac{7}{12}$
(b) $\dfrac{7}{24}$
(c) $\dfrac{5}{12}$
(d) $\dfrac{17}{24}$

Answer 1

Use the formula for the probability of either of the events A or B to occur given as $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$, where $\cup$ is the symbol of union and $\cap$ is the symbol of intersection. Take the expression $P\left( A\cap B \right)$ to the L.H.S and equate the provided value of $P\left( A\cup B \right)+P\left( A\cap B \right)$. Use the given relation $P\left( A \right)=2P\left( B \right)$ and substitute the value of $P\left( B \right)$ in terms of $P\left( A \right)$ to solve for its value.

Complete step-by-step solution:
Here we have been provided with two events A and B with the relations $P\left( A\cup B \right)+P\left( A\cap B \right)=\dfrac{7}{8}$ and $P\left( A \right)=2P\left( B \right)$. We have to find the value of $P\left( A \right)$.
Now, if two events A and B are given then the probability of occurrence of either A or B is given by the formula $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$, where $\cup$ is the symbol of union and $\cap$ is the symbol of intersection. Taking the expression $P\left( A\cap B \right)$ to the L.H.S we get,
$\Rightarrow P\left( A\cup B \right)+P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)$
Substituting the provided value of the expression $P\left( A\cup B \right)+P\left( A\cap B \right)$ we get,
$\Rightarrow P\left( A \right)+P\left( B \right)=\dfrac{7}{8}$ ……. (1)
From the given relation $P\left( A \right)=2P\left( B \right)$ we can write $P\left( B \right)=\dfrac{P\left( A \right)}{2}$, therefore substituting the value of $P\left( B \right)$ in terms of $P\left( A \right)$ in equation (1) we get,
$\begin{aligned} & \Rightarrow P\left( A \right)+\dfrac{P\left( A \right)}{2}=\dfrac{7}{8} \\\ & \Rightarrow \dfrac{3P\left( A \right)}{2}=\dfrac{7}{8} \\\ & \Rightarrow P\left( A \right)=\dfrac{7\times 2}{8\times 3} \\\ & \therefore P\left( A \right)=\dfrac{7}{12} \\\ \end{aligned}$
Hence, option (a) is the correct answer.

Note: Note that the probability value $P\left( A\cup B \right)$ denotes the probability of occurrence of either event A or event B but the probability value $P\left( A\cap B \right)$ denotes the probability of occurrence of both the events A and B simultaneously. In case A and B are independent events, the value of $P\left( A\cap B \right)$ can be found by using the formula $P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)$.