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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 $P(E)<P(F)$, 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: enter image description here

Solution: enter image description here

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    $\begingroup$ It doesn't, clearly. Perhaps you left off some assumptions? $\endgroup$ – lulu Apr 20 at 19:10
  • $\begingroup$ @lulu I've edited the question with the source of my confusion $\endgroup$ – Eagle Apr 20 at 19:15
  • $\begingroup$ @MariaMazur edited the question $\endgroup$ – Eagle Apr 20 at 19:15
  • $\begingroup$ @MariaMazur from where did the first statement come? $\endgroup$ – Eagle Apr 20 at 19:17
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    $\begingroup$ 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". $\endgroup$ – lulu Apr 20 at 19:27
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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

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

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  • $\begingroup$ I've edited the question with the source $\endgroup$ – Eagle Apr 20 at 19:17
  • $\begingroup$ What exactly is the question? I gave a counterexample for the claim you proposed. $\endgroup$ – EpsilonDelta Apr 20 at 19:18
  • $\begingroup$ 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! $\endgroup$ – Eagle Apr 20 at 19:20
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    $\begingroup$ Yes, the solution seems wrong. A weird reasoning for sure. $\endgroup$ – EpsilonDelta Apr 20 at 19:21

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