Question

Let E and F be two independent events. The probability that both E and F happen is ${\displaystyle \frac{1}{12}}$ and the probability that neither E nor F happens is ${\displaystyle \frac{1}{2}}$, then a value of ${\displaystyle \frac{P\left(E\right)}{P\left(F\right)}}$ is
(a) ${\displaystyle \frac{5}{12}}$
(b) ${\displaystyle \frac{1}{3}}$
(c) ${\displaystyle \frac{3}{2}}$
(d) ${\displaystyle \frac{4}{3}}$

Answer 1

Hint: Use the two given probabilities to make two equations. Then, using the formula $P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)$ and $P(E\cap F)=P\left(E\right)\cdot P\left(F\right)$, make two equations and solve them to find the values of $P\left(E\right)$ and $P\left(F\right)$.

“Complete step-by-step answer:”
We know the following facts:
1. The probability that two events A and B happen together is given as $P(A\cap B)$
2. The probability that at least one of the two events A and B happens is given as $P(A\cup B)$
3. The probability that an event E does not happen is given as $1-P\left(E\right)$, if $P\left(E\right)$ is the probability that the event A happens.
Applying the above facts to the statements given in the question:
Probability that E and F happen together is ${\displaystyle \frac{1}{12}}$, which can be written as $P(E\cap F)={\displaystyle \frac{1}{12}}$
The second statement, probability that neither E nor F happen can be understood as the negation of the event that at least one of them happens.
The probability that at least one of E or F happens is given as $P(E\cup F)$.
Hence, the probability of neither E nor F happens is given as $1-P(E\cup F)={\displaystyle \frac{1}{2}}$. Upon rearranging,
$\begin{array}{rl}& \Rightarrow P(E\cup F)=1-{\displaystyle \frac{1}{2}}\\ & \Rightarrow P(E\cup F)={\displaystyle \frac{1}{2}}\end{array}$
Thus, we have two results $P(E\cap F)={\displaystyle \frac{1}{12}}$ and $P(E\cup F)={\displaystyle \frac{1}{2}}$.
We know that $P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)$.
Substituting the value of $P(E\cup F)$ and $P(E\cap F)$ in the above formula, we get
$$\begin{array}{rl}& {\displaystyle \frac{1}{2}}=P\left(E\right)+P\left(F\right)-{\displaystyle \frac{1}{12}}\\ & \Rightarrow P\left(E\right)+P\left(F\right)={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{12}}\\ & \Rightarrow P\left(E\right)+P\left(F\right)={\displaystyle \frac{7}{12}}\dots \left(1\right)\end{array}$$
Also, since the events E and F are independent, $P(E\cap F)=P\left(E\right)\cdot P\left(F\right)$
Thus, $P\left(E\right)\cdot P\left(F\right)={\displaystyle \frac{1}{12}}\dots \left(2\right)$
To solve the equations (1) and (2) to find $P\left(E\right)$ and $P\left(F\right)$, we can use the relation $a-b=\sqrt{{(a+b)}^2-4ab}$
In this equation, $a=P\left(E\right)$ and $b=P\left(F\right)$
$$\stackrel{2}{(P\left(E\right)-P\left(F\right))}={(P\left(E\right)+P\left(F\right))}^2-4P\left(E\right)\cdot P\left(F\right)$$
Substituting values from equations (1) and (2),
$$\begin{array}{rl}& \Rightarrow \stackrel{2}{(P\left(E\right)-P\left(F\right))}={\left({\displaystyle \frac{7}{12}}\right)}^2-4\left({\displaystyle \frac{1}{12}}\right)\\ & \Rightarrow \stackrel{2}{(P\left(E\right)-P\left(F\right))}=\left({\displaystyle \frac{49}{144}}\right)-\left({\displaystyle \frac{1}{3}}\right)\\ & \Rightarrow \stackrel{2}{(P\left(E\right)-P\left(F\right))}={\displaystyle \frac{1}{144}}\\ & \Rightarrow P\left(E\right)-P\left(F\right)={\displaystyle \frac{1}{12}}\dots \left(3\right)\end{array}$$
Adding equations (1) and (3),
$$2\cdot P\left(E\right)={\displaystyle \frac{8}{12}}$$
$$\Rightarrow P\left(E\right)={\displaystyle \frac{4}{12}}={\displaystyle \frac{1}{3}}$$
Subtracting equation (3) from equation (1), we get
$$2\cdot P\left(F\right)={\displaystyle \frac{6}{12}}$$
$$\Rightarrow P\left(F\right)={\displaystyle \frac{3}{12}}={\displaystyle \frac{1}{4}}$$
Thus, the required value, ${\displaystyle \frac{P\left(E\right)}{P\left(F\right)}}={\displaystyle \frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{4}}}}={\displaystyle \frac{4}{3}}$
Therefore, the correct answer is option (d).

Note: The formula used here, $P(E\cap F)=P\left(E\right)\cdot P\left(F\right)$ is only valid if the two events E and F are independent of each other (given in the question). Otherwise this formula is not applicable, and then using this formula would result in an incorrect answer.