Animated Solution for Mathematics - Probability: If $E$ and $F$ are events with $P(E)\le P(F)$ and $P(E\cap F)>0$, then

Probability vs. Logic

Hello, future engineers! Today, we are tackling a problem that often trips up even the brightest minds in the JEE Advanced arena. It is a question that tests your ability to distinguish between numerical probability and logical set theory.

Deconstructing the Conditions

We are given two events, $E$ and $F$, in a sample space $S$. We have two conditions: $P(E)\le P(F)$ and $P(E\cap F)>0$.

Your intuition might suggest that if $E$ is smaller than $F$ and they overlap, $E$ must be inside $F$. However, in rigorous mathematics, intuition must be verified. The condition $P(E)\le P(F)$ is purely a statement about the "size" or "weight" of the events, telling us nothing about their relative positions.

The second condition, $P(E\cap F)>0$, simply means that the events are not disjoint and share at least some common outcomes. It does not inherently force $E$ to be a subset of $F$.

The Logic of Implication

In set theory, the statement "occurrence of $E\Rightarrow$ occurrence of $F$" is logically equivalent to $E\subseteq F$. Similarly, "non-occurrence of $E\Rightarrow$ non-occurrence of $F$" is the contrapositive of $F\Rightarrow E$, which implies $F\subseteq E$.

The core question is whether these conditions force $E\subseteq F$ or $F\subseteq E$. If we can find a single scenario where neither is true, we have our answer.

The Power of the Counterexample

This is the ultimate JEE weapon: the counterexample. Let us define a sample space $S=\{1,2,3,4,5\}$ where each outcome is equally likely, such that $P(\{i\})=\frac{1}{5}$.

Now, let us define the sets $E=\{1,2\}$ and $F=\{2,3,4\}$.

First, we verify the conditions: 1. $P(E)=\frac{2}{5}$ and $P(F)=\frac{3}{5}$. Since $\frac{2}{5}\le \frac{3}{5}$, the first condition is satisfied. 2. The intersection $E\cap F=\{2\}$, so $P(E\cap F)=\frac{1}{5}$. Since $\frac{1}{5}>0$, the second condition is also satisfied.

Now, we test the implications: - Is $E\subseteq F$? No, because $1\in E$ but $1 otin F$. Thus, $E\Rightarrow F$ is false. - Is $F\subseteq E$? No, because $3\in F$ but $3 otin E$. Thus, $F\Rightarrow E$ is false. - Since $F ot\subseteq E$, the contrapositive $ eg E \Rightarrow eg F$ is also false.

The Verdict

We have successfully constructed a scenario where all given conditions hold, yet none of the implications are true. This proves that $P(E)\le P(F)$ and $P(E\cap F)>0$ are insufficient to force any of the logical relationships.

The correct answer is (d): none of the above implications holds. Remember, in probability, always look for the counterexample before you commit to an implication.