Answer

Verified

450.9k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Three liquids are given to you One is hydrochloric class 11 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE