• Jul 12th 2012, 12:58 PM
Rikimaru
Can someone check the following for me please? I'm confused by the wording of the question :(.

Quote:

Consider the experiment of drawing 3 tickets without replacement from a box of tickets labeled {1, 2, 3, ... , 10}.
Let A be the event that at least one of the tickets drawn is labeled with an even number,
Let B be the event that at least one of the tickets drawn is labeled with a prime number,
Let C be the event that at least one of the tickets drawn is labeled with a number larger than 9, and
Let D be the event that the sum of the numbers on the tickets drawn is at least 25.

2. Are A and B mutually exclusive?

No (this would work since drawing a 2 would satisfies the even and prime factor)

3. Are A and Bc mutually exclusive?
No (unsure)

4. Are C and Dc mutually exclusive?
No (unsure)

What does Bc and Dc stand for here? So basically Bc would mean Not B... would it be fair to say that Bc = A∪C∪D?... so Dc = A∪B∪C

5. Which of the following is/are true (select all that apply)?
A. A implies B (even doesn't have to be prime)
B. B implies A (prime doesn't have to be even)
C. C implies D (having a 10 doesn't imply that the total sum will be 25; it could just be 10, 1, 2)
D. D implies C ***
E. none of the above
>>> D implies C since the sum total of 25 requires at least 10 to be in one of the selection. (10 + 9 + 6 or 10 + 8 + 7 are the only 2 solutions)... 9 + 8 + 7 is only 24, thus forcing D to include 10, thus satisfying C.

Thanks!
• Jul 12th 2012, 03:27 PM
mac01021
2. Right, but not only for the reason cited, since drawing a 4 and a 7 together would also satisfy both events.

3 & 4. I believe you also have the right answers here. See wikipedia [http://en.wikipedia.org/wiki/Complement_(set_theory)] for an introduction to this notation (The c should be in superscript, though.)

5. It seems like you don't actually need help with any of this. Gook work.

If there is any part of the wording other than the bizarre complement notation that confuses you, point out what it is so people can try to help.
• Jul 12th 2012, 03:49 PM
Rikimaru
Anyways thanks! I ended up getting 100% on the assignment. But I just want to clear #3 and #4 since I'm still a bit confused...

Although I know that they're complement, I'm still confused if I interpreted it correctly here...

Would Bc mean that: Not B (not prime)... so it must be A (even), C (10), D (sum: 25)?
--> figured this combo would work:
10 (A & C), 9, 6 (A) = 25 (D); no prime

Would Dc mean that: Not D (not total sum 25)... so it must be A (even), B (prime), C (10),
--> figured this combo would work (probably many more since D restricted the possible solutions to 2):
10 (A & C), 7 (B), 2 (random); sum total not 25
• Jul 12th 2012, 04:20 PM
mac01021
Quote:

Originally Posted by Rikimaru

Would Bc mean that: Not B (not prime)... so it must be A (even), C (10), D (sum: 25)?
--> figured this combo would work:
10 (A & C), 9, 6 (A) = 25 (D); no prime

I'm having trouble understanding exactly what you mean here. I suspect that you do understand correctly but, just to be sure:

[1] $B^c$ includes any drawing that contains none of 2,3,5,7 (all the primes in the box).
[2] $A$ includes any drawing containing at least one of 2,4,6,8,10.

So if you there is a possible drawing which contains one or more of the tickets listed in [2], but none of the tickets listed in [1], then $A \cup B^c \neq \emptyset$ which means the two events are not mutually exclusive.
You found one such possible drawing, (10,9,6), and therefore came to the right conclusion .

Quote:

Originally Posted by Rikimaru
Would Dc mean that: Not D (not total sum 25)... so it must be A (even), B (prime), C (10),
--> figured this combo would work (probably many more since D restricted the possible solutions to 2):
10 (A & C), 7 (B), 2 (random); sum total not 25

Similar reasoning applies here.