Most & Some

Comments

• 2 Likes

  barista17 Member

  January 2021 112 karma

  Here is an example using letters:

  AAAAAAAAAA
  BBBBBBBBBBBBBBBBBBBBBBBBBB
  .....................................CCCCCCCCCCCCC

  All "A" is B. Most "B" is "C." There is not at least one (some) A that is a C. That is, there is no "A" that is "C." So, no valid inference can be made.
• hellome Core Member

  January 2021 edited January 2021 44 karma

  @barista17 said:
  Here is an example using letters:

  AAAAAAAAAA
  BBBBBBBBBBBBBBBBBBBBBBBBBB
  .....................................CCCCCCCCCCCCC

  All "A" is B. Most "B" is "C." There is not at least one (some) A that is a C. That is, there is no "A" that is "C." So, no valid inference can be made.

  Thank you!!!

  I think I get it. So just B doesn't mean A because the correct contrapositive of A----->B is /B----->/A right? But satisfying the necessary condition doesn't lead anywhere.....I think....

  I think my mistake is assuming every B is A
• 1 Like

  McBeck418 Member

  January 2021 500 karma

  You’re correct. If you affirm the necessary condition (B) you don’t know anything else. You can’t assume that if you have a B you also have an A.
• 2 Likes

  whatsmyname Member

  January 2021 606 karma

  Use a venn diagram when trying to break down some and most concepts.

  also draw elements on a piece of paper and try to make sense of them like @barista17 did!
• 1 Like

  FaviPapi-1-1 Member

  January 2021 edited January 2021 313 karma

  Diagram 1: A -----> B ---most---> C

  From Diagram 1, we cannot come up with any proper inferences. The reason to that is the following: we do not know how big the B group is...

  Let's say there are 10 people in the A group, and all those 10 people in such group are in the B group. How big is the B group? Let's assume it's 100 people in such group; 10 of those 100 people are from the A group.

  Now, most Bs are in C. It could that 90 people of those 100 people of group B that were not from group A are in C. If that is the case, then we cannot conclude that some As are Cs since none of the As were in C...

  I hope this helps; good luck.
• hellome Core Member

  February 2021 44 karma

  Super helpful. Thank you so much!
• 1 Like

  Sailor Moon LSAT Member ⭐

  February 2021 200 karma

  Also, just know that if a universal quantifier (all, every, any etc) goes first, followed by an existential quantifier (some, most) you can't draw an inference. I remember it by thinking of the some/most words as a road block, but when the straight up ---> arrow is second, I think of it as a rope that allows you to pull through the elements to make an inference.
• hellome Core Member

  February 2021 44 karma

  That's brilliant! Thank you so, so much @"Merly Goodleaf"!