Lesson 13 – Drill Set 2

    • July 9, 2017 at 10:01 am #3151

      Hey Mike,

      Can you provide a walkthrough of Set 2, problem 1 in lesson 13: Conditional Rules?

      The game is as follows:

      A young girl will pack some of her seven dolls-F, G, H, I, K, L, and
      M-to take on a trip. The following rules apply:

      She will take K or G, but not both.
      She will take H or L, but not both.
      If she takes M, she will take F.
      If she doesn’t take I , she won’t take L.
      If she takes K, she will not take F.

      In particular, I’m having difficulty understanding why the answer for question #2 is C.


    • July 10, 2017 at 3:11 pm #3154
      Mike Kim

      Hey Dylan —

      Sure thing —

      Let’s first walk through the rules, then that Q —

      The first rule, which I notated by putting K and G into the T diagram with the double arrow, tells us that of K and G, exactly 1 will be taken and exactly 1 will not be. Per this rule, if we find out that either K or G is taken, we know for sure that the other is not, and if we find out that K or G is not taken, we know for sure that the other one was.

      Similarly, the second rule tells us that of H and L, exactly one will be taken and one will not, and we can make the same types of inferences.

      The third rule tells us that if M is taken, then F must be as well. We can infer from this that if F is not taken, M will not be taken (the contrapositive).

      The fourth rule tells us that if I is not taken, L won’t be taken. We can infer from this the contrapositive that if L is taken, I must have been taken.

      Finally, the last rule tells us that if K is taken, F will not be. So, if F is taken, we know K must not have been.

      So, I hope that’s clear and I hope it makes sense how the notations I used on page 187 represent those rules.

      For Q # 2, we are asked to find an answer that, basically, allows us to know completely and exactly which dolls were taken and which ones were not. The T diagram work to the right of that q corresponds to each of the answer choices —

      Walking through it together —

      (A) allows us to put F and H into the “in” column.

      Per the last rule, we know that F in -> K out.

      Per the first rule, K out -> G in.

      Per the second rule, H in -> L out.

      The inference chain stops at that point, so, we don’t know whether M is in or out, and we don’t know whether I is in or out.

      (B) allows us to put F in the in column and H into the out column.

      If F is in, like last time, we know K is out, and thus G is in.

      If H is out, per the second rule L is in. If L is in, per the fourth rule I is in.

      Our chain ends there — we don’t know whether M is in or out and so (B) is not the correct answer.

      (C) allows us to put M in and H out.

      If M is in, per the third rule, F is in.

      If F is in, just as before, we know that K is out and G is in.

      And, per the second rule, if H is out, L is in. If L is in, per the fourth rule, I must be in.

      So, answer choice (C) allows us to determine for all 7 elements whether they were taken or not.

      I’ll stop there but the same process can help us see that (D) and (E) do not allow us to place all elements.

      I hope that helps clear things up — if you have any follow-up or need anything else let me know —


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