Re: Statistics Help Please

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.

Re: Statistics Help Please

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

Re: Statistics Help Please

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] $\displaystyle B^c$ includes any drawing that contains none of 2,3,5,7 (all the primes in the box).

[2]$\displaystyle A$ 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 $\displaystyle A \cup B^c \neq \emptyset$ 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.