Question

If $A$ and $B$ are independent events of a random experiment, such that $P(A \cap B) = \dfrac{1}{6}$, $P(A' \cap B') = \dfrac{1}{3}$ Then $P(A)$ is equal to
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{5}{7}$
D. $\dfrac{2}{3}$

Answer 1

The question is related to the probability topic. We have to determine the value of probability of event A. By using the properties and formulas of probability we obtain the quadratic equation in the terms of P(A) then by using the formula of root of quadratic equation we are determining the value of P(A).

Formula used:
1. The property of probability $P(A' \cap B') = P((A \cup B)')$
2. If the equation is of the form $a{x^2} + bx + c = 0$ then the root of a quadratic equation is determined by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
Here the two events are independent events.
Given $P(A \cap B) = \dfrac{1}{6}$ and $P(A' \cap B') = \dfrac{1}{3}$
As we know that $P(A' \cap B') = P((A \cup B)')$
So we have
$\Rightarrow P((A \cup B)') = \dfrac{1}{3}$
By the formula $P((A \cup B)') = 1 - P(A \cup B)$
On substituting the values we have
$\Rightarrow \dfrac{1}{3} = 1 - P(A \cup B)$

On shifting the terms we have
$\Rightarrow P(A \cup B) = 1 - \dfrac{1}{3}$
On simplifying we have
$\Rightarrow P(A \cup B) = \dfrac{2}{3}$----- (1)
As we know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. On substituting the known values
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
On substituting the known values we have
$\Rightarrow \dfrac{2}{3} = P(A) + P(B) - \dfrac{1}{6}$
Take $\dfrac{1}{6}$ to LHS
$\Rightarrow \dfrac{2}{3} + \dfrac{1}{6} = P(A) + P(B)$
On simplification we have
$\Rightarrow P(A) + P(B) = \dfrac{5}{6}$ ----- (1)

Since the two events are independent events, so we have $P(A \cap B) = P(A).P(B)$. Substituting the known values to above formula we get
$\Rightarrow P(A).P(B) = \dfrac{1}{6}$ -------- (2)
On considering the equation (1) and write in terms of P(B) we have
$\Rightarrow P(B) = \dfrac{5}{6} - P(A)$ ------(3)
On substituting the equation (3) in the equation (2) we have
$\Rightarrow P(A).\left( {\dfrac{5}{6} - P(A)} \right) = \dfrac{1}{6}$
On multiplying we get
$\Rightarrow \dfrac{5}{6}P(A) - {\left( {P(A)} \right)^2} = \dfrac{1}{6}$
$\Rightarrow {\left( {P(A)} \right)^2} - \dfrac{5}{6}P(A) + \dfrac{1}{6} = 0$
$\Rightarrow 6{\left( {P(A)} \right)^2} - 5P(A) + 1 = 0$

The above equation is in the form of a quadratic equation in terms of P(A). The above quadratic equation will be written as
$\Rightarrow 6{\left( {P(A)} \right)^2} - 3P(A) - 2P(A) + 1 = 0$
$\Rightarrow 3P(A)\left( {2P(A) - 1} \right) - 1\left( {2P(A) - 1} \right) = 0$
$\Rightarrow \left( {2P(A) - 1} \right)\left( {3P(A) - 1} \right) = 0$
Now both terms are equated to 0 so we have
$\Rightarrow \left( {2P(A) - 1} \right) = 0$ or $\left( {3P(A) - 1} \right) = 0$
On simplification
$\therefore P(A) = \dfrac{1}{2}$ or $P(A) = \dfrac{1}{3}$
The P(A) has two values, when we compare the obtained answer to the given options

Hence option B is the correct answer.

Note: When the equation is in the form of a quadratic equation. The roots can be solved by using the sum product rule. This defines as for the general quadratic equation $a{x^2} + bx + c$, the product of $a{x^2}$ and c is equal to the sum of $bx$ of the equation by using this we can form the equation and by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we can determine the roots for the equation.