Question

If E and F are events with $P\left( E \right)\le P\left( F \right)$ and $P\left( E\bigcap F \right) > 0$, then
$\begin{align}
  & \left[ a \right]\text{Occurence of E }\Rightarrow \text{ Occurence of F} \\
 & \left[ b \right]\text{ Occurence of F }\Rightarrow \text{ Occurence of E} \\
 & \left[ c \right]\text{ Non-occurrence of E }\Rightarrow \text{Non-occurrence of F} \\
 & \left[ d \right]\text{ None of these} \\
\end{align}$

Answer 1

Hint: Use the fact that if the occurrence of an event E implies the occurrence of an event F, then $P\left( F/E \right)=1$ Consider the event of throwing a die and define event E as the event of getting a number which is a multiple of 3 and F as the event of getting a number which is a multiple of 2. Hence determine which of the options is correct and which of the options is incorrect.

Complete step-by-step solution -
We will check option wise which of the options is correct and which of the options is incorrect.
Option [a]:
Consider the random experiment of throwing a die.
Let E be the event that the number obtained is a multiple of 3
Let F be the event that the number obtained is a multiple of 2.
Hence, we have
$E=\left\{ 3,6 \right\}$
Hence, we have
$P\left( E \right)=\dfrac{2}{6}=\dfrac{1}{3}$
Also, we have
$F=\left\{ 2,4,6 \right\}$
Hence, we have
$P\left( F \right)=\dfrac{3}{6}=\dfrac{1}{2}$
Hence, we have $P\left( E \right)\le P\left( F \right),P\left( E\bigcap F \right)=P\left( \left\{ 6 \right\} \right) > 0$
Hence all the conditions of the question are satisfied.
Now, we have $P\left( F/E \right)=\dfrac{P\left( E\bigcap F \right)}{P\left( E \right)}=\dfrac{\dfrac{1}{6}}{\dfrac{1}{3}}=\dfrac{1}{2}\ne 1$
Hence, the occurrence of E does not imply the occurrence of F.
Hence option [a] is incorrect.
Option [b]
Consider the random experiment of throwing a die.
Let E be the event that the number obtained is a multiple of 3
Let F be the event that the number obtained is a multiple of 2.
Hence, we have
$E=\left\{ 3,6 \right\}$
Hence, we have
$P\left( E \right)=\dfrac{2}{6}=\dfrac{1}{3}$
Also, we have
$F=\left\{ 2,4,6 \right\}$
Hence, we have
$P\left( F \right)=\dfrac{3}{6}=\dfrac{1}{2}$
Hence, we have $P\left( E \right)\le P\left( F \right),P\left( E\bigcap F \right)=P\left( \left\{ 6 \right\} \right) > 0$
Hence all the conditions of the question are satisfied.
Now, we have $P\left( E/F \right)=\dfrac{P\left( E\bigcap F \right)}{P\left( F \right)}=\dfrac{\dfrac{1}{6}}{\dfrac{1}{2}}=\dfrac{1}{3}\ne 1$
Hence, the occurrence of F does not imply the occurrence of E.
Hence option [b] is incorrect.
Option [c]:
Consider the random experiment of throwing a die.
Let E be the event that the number obtained is a multiple of 3
Let F be the event that the number obtained is a multiple of 2.
Hence, we have
$E=\left\{ 3,6 \right\}$
Hence, we have
$P\left( E \right)=\dfrac{2}{6}=\dfrac{1}{3}$
Also, we have
$F=\left\{ 2,4,6 \right\}$
Hence, we have
$P\left( F \right)=\dfrac{3}{6}=\dfrac{1}{2}$
Also, we have
$E\bigcap F=\left\{ 6 \right\}\Rightarrow P\left( E\bigcap F \right)=\dfrac{1}{6}$ and $E\bigcup F=\left\{ 2,3,4,5,6 \right\}\Rightarrow P\left( E\bigcup F \right)=\dfrac{5}{6}$
Hence, we have $P\left( E \right)\le P\left( F \right),P\left( E\bigcap F \right)=P\left( \left\{ 6 \right\} \right) > 0$
Hence all the conditions of the question are satisfied.
Now, we have $P\left( {{F}^{c}}/{{E}^{c}} \right)=\dfrac{P\left( {{E}^{c}}\bigcap {{F}^{c}} \right)}{P\left( {{E}^{c}} \right)}=\dfrac{P\left( {{\left( E\bigcup F \right)}^{c}} \right)}{1-\dfrac{1}{3}}=\dfrac{1-\dfrac{5}{6}}{1-\dfrac{1}{3}}=\dfrac{1}{4} \ne 1$
Hence, the non-occurrence of E does not imply the non-occurrence of F.
Hence option [c] is incorrect.
Hence option [d] is correct.

Note: In these types of questions the most common mistakes made by the student is that they give an example in which one of the options holds correct and the rest others do not and claim that the option is correct. This process is clearly incorrect since if we want to prove that some mathematical statement is correct, we need to come up with formal proof. If we want to prove that some mathematical statement is incorrect, we need to come up with a counterexample as is done in the above solution.