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(a) $\dfrac{5}{12}$

(b) $\dfrac{1}{3}$

(c) $\dfrac{3}{2}$

(d) $\dfrac{4}{3}$

Answer

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

2. The probability that at least one of the two events A and B happens is given as $P\left( A\cup B \right)$

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 $\dfrac{1}{12}$, which can be written as $P\left( E\cap F \right)=\dfrac{1}{12}$

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 $P\left( E\cup F \right)$.

Hence, the probability of neither E nor F happens is given as $1-P\left( E\cup F \right)=\dfrac{1}{2}$. Upon rearranging,

$\begin{align}

& \Rightarrow P\left( E\cup F \right)=1-\dfrac{1}{2} \\

& \Rightarrow P\left( E\cup F \right)=\dfrac{1}{2} \\

\end{align}$

Thus, we have two results $P\left( E\cap F \right)=\dfrac{1}{12}$ and $P\left( E\cup F \right)=\dfrac{1}{2}$.

We know that $P\left( E\cup F \right)=P\left( E \right)+P\left( F \right)-P\left( E\cap F \right)$.

Substituting the value of $P\left( E\cup F \right)$ and $P\left( E\cap F \right)$ in the above formula, we get

\[\begin{align}

& \dfrac{1}{2}=P\left( E \right)+P\left( F \right)-\dfrac{1}{12} \\

& \Rightarrow P\left( E \right)+P\left( F \right)=\dfrac{1}{2}+\dfrac{1}{12} \\

& \Rightarrow P\left( E \right)+P\left( F \right)=\dfrac{7}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\

\end{align}\]

Also, since the events E and F are independent, $P\left( E\cap F \right)=P\left( E \right)\cdot P\left( F \right)$

Thus, $P\left( E \right)\cdot P\left( F \right)=\dfrac{1}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right)$

To solve the equations (1) and (2) to find $P\left( E \right)$ and $P\left( F \right)$, we can use the relation $a-b=\sqrt{{{\left( a+b \right)}^{2}}-4ab}$

In this equation, $a=P\left( E \right)$ and $b=P\left( F \right)$

\[\mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}={{\left( P\left( E \right)+P\left( F \right) \right)}^{2}}-4P\left( E \right)\cdot P\left( F \right)\]

Substituting values from equations (1) and (2),

\[\begin{align}

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}={{\left( \dfrac{7}{12} \right)}^{2}}-4\left( \dfrac{1}{12} \right) \\

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}=\left( \dfrac{49}{144} \right)-\left( \dfrac{1}{3} \right) \\

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}=\dfrac{1}{144} \\

& \Rightarrow P\left( E \right)-P\left( F \right)=\dfrac{1}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 3 \right) \\

\end{align}\]

Adding equations (1) and (3),

\[2\cdot P\left( E \right)=\dfrac{8}{12}\]

\[\Rightarrow P\left( E \right)=\dfrac{4}{12}=\dfrac{1}{3}\]

Subtracting equation (3) from equation (1), we get

\[2\cdot P\left( F \right)=\dfrac{6}{12}\]

\[\Rightarrow P\left( F \right)=\dfrac{3}{12}=\dfrac{1}{4}\]

Thus, the required value, $\dfrac{P\left( E \right)}{P\left( F \right)}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}=\dfrac{4}{3}$

Therefore, the correct answer is option (d).

Note: The formula used here, $P\left( E\cap F \right)=P\left( E \right)\cdot P\left( F \right)$ 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.

“Complete step-by-step answer:”

We know the following facts:

1. The probability that two events A and B happen together is given as $P\left( A\cap B \right)$

2. The probability that at least one of the two events A and B happens is given as $P\left( A\cup B \right)$

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 $\dfrac{1}{12}$, which can be written as $P\left( E\cap F \right)=\dfrac{1}{12}$

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 $P\left( E\cup F \right)$.

Hence, the probability of neither E nor F happens is given as $1-P\left( E\cup F \right)=\dfrac{1}{2}$. Upon rearranging,

$\begin{align}

& \Rightarrow P\left( E\cup F \right)=1-\dfrac{1}{2} \\

& \Rightarrow P\left( E\cup F \right)=\dfrac{1}{2} \\

\end{align}$

Thus, we have two results $P\left( E\cap F \right)=\dfrac{1}{12}$ and $P\left( E\cup F \right)=\dfrac{1}{2}$.

We know that $P\left( E\cup F \right)=P\left( E \right)+P\left( F \right)-P\left( E\cap F \right)$.

Substituting the value of $P\left( E\cup F \right)$ and $P\left( E\cap F \right)$ in the above formula, we get

\[\begin{align}

& \dfrac{1}{2}=P\left( E \right)+P\left( F \right)-\dfrac{1}{12} \\

& \Rightarrow P\left( E \right)+P\left( F \right)=\dfrac{1}{2}+\dfrac{1}{12} \\

& \Rightarrow P\left( E \right)+P\left( F \right)=\dfrac{7}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\

\end{align}\]

Also, since the events E and F are independent, $P\left( E\cap F \right)=P\left( E \right)\cdot P\left( F \right)$

Thus, $P\left( E \right)\cdot P\left( F \right)=\dfrac{1}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right)$

To solve the equations (1) and (2) to find $P\left( E \right)$ and $P\left( F \right)$, we can use the relation $a-b=\sqrt{{{\left( a+b \right)}^{2}}-4ab}$

In this equation, $a=P\left( E \right)$ and $b=P\left( F \right)$

\[\mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}={{\left( P\left( E \right)+P\left( F \right) \right)}^{2}}-4P\left( E \right)\cdot P\left( F \right)\]

Substituting values from equations (1) and (2),

\[\begin{align}

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}={{\left( \dfrac{7}{12} \right)}^{2}}-4\left( \dfrac{1}{12} \right) \\

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}=\left( \dfrac{49}{144} \right)-\left( \dfrac{1}{3} \right) \\

& \Rightarrow \mathop{\left( P\left( E \right)-P\left( F \right) \right)}^{2}=\dfrac{1}{144} \\

& \Rightarrow P\left( E \right)-P\left( F \right)=\dfrac{1}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 3 \right) \\

\end{align}\]

Adding equations (1) and (3),

\[2\cdot P\left( E \right)=\dfrac{8}{12}\]

\[\Rightarrow P\left( E \right)=\dfrac{4}{12}=\dfrac{1}{3}\]

Subtracting equation (3) from equation (1), we get

\[2\cdot P\left( F \right)=\dfrac{6}{12}\]

\[\Rightarrow P\left( F \right)=\dfrac{3}{12}=\dfrac{1}{4}\]

Thus, the required value, $\dfrac{P\left( E \right)}{P\left( F \right)}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}=\dfrac{4}{3}$

Therefore, the correct answer is option (d).

Note: The formula used here, $P\left( E\cap F \right)=P\left( E \right)\cdot P\left( F \right)$ 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.