# Lesson 13 – Drill Set 2

• July 9, 2017 at 10:01 am #3151
Dylan
Participant

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.

Thanks!

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

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 —

Mike

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