##### Quiz 3: First Level C Question
AZobi
You have three pants and four shirts. Exactly one shirt is blue and on pair of pants is blue. You randomly select and outfit. What is the probability that at least one clothing item you choose is blue? I decided to post the remaining questions on the discussion page so everyone can benefit from the collaborations.
AZobi: Feb. 6, 2015, 8:06 p.m.
My Answer: # Of different outfits is 12 given that there are three pants and four shirts. From those different outfits, 6 outfits contain at least one article of blue clothing (one of them contains two articles of blue clothing). The probability will then be 6/12 or 1/2.
AZobi: Feb. 6, 2015, 8:07 p.m.
Can someone check to make sure I did it correctly? I am often prone to making errors and over-looking some details.
ShivaniGillon: Feb. 6, 2015, 11:40 p.m.
I also got the same answer :)
weisbart: Feb. 7, 2015, 11:36 a.m.
Good job! Thanks also for posting the questions, AZobi. I really do want you guys to all work together. It helps everyone, including me. You guys learn from each other and learn by explaining and I can see what your thinking is and how to improve my explanations. Everyone does better in the end!
weisbart: Feb. 7, 2015, 11:38 a.m.
Here is a question: How did you calculate that the numerator is 12? Of course, you could just write out all possibilities, but what if the numbers were very large, say 100 shirts and 200 pants?
AZobi: Feb. 7, 2015, 12:32 p.m.
No problem! :) And I calculated 12 by multiplying the number of shirts by the number of pants to determine the total number of outfits. I did this because of the Law of Products. To answer your question, there will be 20,000 possible outfits after multiplying 100 by 200.
weisbart: Feb. 8, 2015, 10:20 a.m.
Oh...wait I'm sorry. I asked the wrong question. How did you get the 6? You're right of course, but I'm curious if you counted out the possibilities or if you have a way that could be extended to larger numbers.
AZobi: Feb. 8, 2015, 3:01 p.m.
Well, I did count out the ways. After discussing it with rrakha and marmat1, we also solved it doing 1- [(2 choose 1)(3 choose 1)/(4 choose 1)(3 choose 1)]. In other words, 1 minus the ways of choosing an outfit that does not have a blue article of clothing.
ANguyen: Feb. 8, 2015, 5:01 p.m.
how would you count if you wanted to do larger numbers?
AZobi: Feb. 8, 2015, 5:48 p.m.
Would it be 1- (99 choose 1) (199 choose 1) all over (200)(100)? My reasoning would be that the chances of choosing at least one blue item would be one minus the ways of choosing anything but blue.
weisbart: Feb. 8, 2015, 8:44 p.m.
Working with the complement is a good idea. I suppose I just want to point out that 6 comes up in the following way: 1 blue shirt paired with 2 non blue pants, 3 non blue shirts paired with 1 blue pair of pants, finally, 1 blue shirt and 1 blue pair of pants. This gives $1\times 2 + 3\times 1 + 1\times 1 = 6$ appropriate outfits. I just want to point out that it is very easy to over count the doubled blue outfit.