# Why does $P(E) < P(F)$ imply that $E \subseteq F$?

Why does $$P(E) < P(F)$$ mean that $$E \subseteq F$$ ?

My reasoning (using Venn diagrams):

It is seen clearly in the below picture that even if $$P(E), there is still some region in E that is not a part of F, So why is $$E \subseteq F$$ true?

Source if this information

Solution for a question from JEE Advanced 1998

Question:

Solution:

• It doesn't, clearly. Perhaps you left off some assumptions? – lulu Apr 20 at 19:10
• @lulu I've edited the question with the source of my confusion – Eagle Apr 20 at 19:15
• @MariaMazur edited the question – Eagle Apr 20 at 19:15
• @MariaMazur from where did the first statement come? – Eagle Apr 20 at 19:17
• Well, if that's all they wrote then I agree, it's incomprehensible. Your approach is entirely correct. Just construct counterexamples to each of the other options and then you are left with "none of the above". – lulu Apr 20 at 19:27

It is false.

Consider cossing two coins. Write $$H$$ when we get a head and $$T$$ when we get tail. Then

$$\Omega := \{(H,H),(H,T),(T,T),(T,H)\}$$

Then $$P(\{(H,H)\}) = 1/4 < P(\{(T,T),(T,H)\}) = 1/2$$

yet

$$\{(H,H)\} \not\subseteq \{(T,T),(T,H)\}$$

• I've edited the question with the source – Eagle Apr 20 at 19:17
• What exactly is the question? I gave a counterexample for the claim you proposed. – EpsilonDelta Apr 20 at 19:18
• I was confused as the statement was given in a solution. Even I wasn't sure if it was true. Now, MariaMazur had clarified in the comments that the solution is wrong. Thanks for your time! – Eagle Apr 20 at 19:20
• Yes, the solution seems wrong. A weird reasoning for sure. – EpsilonDelta Apr 20 at 19:21