8 television sets, 2 are black and white, 6 others are color. If you had to pick two tv sets out of the 8 randomly, what is the probability that AT LEAST 1 will be black and white?

OMG IS THIS A HOMEWORK QUESTION???

haha no. a friend of mine is studying for GMAT (which I have to as well) and is unable to answer that.

13/28ths is what I've got figured out. It's a combination question, so when picking 2 out of 8, you should get 28 different possibilties of pairings, of which 13 should contain at least one black & white set, since you've got 1 black & white paired with each of the 6 colors twice plus the two black and white TVs together...6+6+1 = 13.

do you mean 13/26?.. in which i case i follow you. thanks btw.

No there's 28 combinations. Write down what you have...black & white 1, black & white 2, color 1, color 2, color 3, color 4, color 5, color 6. Now write down all possible pairings.

1. Black & white 1, black & white 2
2. Black & white 1, color 1
3. Black & white 1, color 2
4. Black & white 1, color 3
5. Black & white 1, color 4
6. Black & white 1, color 5
7. Black & white 1, color 6
8. Black & white 2, color 1
9. Black & white 2, color 2
10. Black & white 2, color 3
11. Black & white 2, color 4
12. Black & white 2, color 5
13. Black & white 2, color 6
14. color 1, color 2
15. color 1, color 3
16. color 1, color 4
17. color 1, color 5
18. color 1, color 6
19. color 2, color 3
20. color 2, color 4
21. color 2, color 5
22. color 2, color 6
23. color 3, color 4
24. color 3, color 5
25. color 3, color 6
26. color 4, color 5
27. color 4, color 6
28. color 5, color 6

28 possible pairs (none repeated and order of choice doesn't count), 13 of which contain at least 1 Black & White set. 13 out of 28.

ahh okay. makes sense...i didn't think there were so many combos. merci.

thread can be closed before phix starts posting crappy gifs.

Well I started out on an online combinations calculator, he should know those equations for combinations and permutations for the GMAT. Ya know, the ones involving factorials. I'm just demonstrating that the answer is valid, I think the method should be his aim.

I did stats during the first year of my BBA course...I plan on starting for GMAT soon and brushing up on the knowledge.

Are you familiar with GMAT? My exam's on Jan 15..think it's enough time? Any tips you can give me...

I only know of the GMAT, I don't have any real advice for it or plan to take it...

fair enough.

I may be wrong, cause i never took this crap in school. My pea sized brain says " theres 8 tvs total, 2 of which are black and white, thats 25 percent" so theres a 25 percent chance that one picked and random will be black and white.

WAIT A MOTHERFUCKING MINUTE?!?!?! A MOTHERFUCKING HOMEWORK QUESTION?! WHERE IS THE HATE?! THIS IS BULLSHIT!! ASKING FOR HELP ON AN INTERNET FORUM IS AMORAL AND EVIDENTALLY UP UNTIL TODAY WE DID NOT CONDONE IT.

well moving on
The value you are looking for is found by first finding out how many possible pairings there are 8C2 = 28
xCy= (x!)/((y-x)!*(y!))
Then like clark said it is semi obvious with a little intuitive thinking that there are 13 combinations that involve the tv being B&W. Theres probably an equation to solve a little quicker but not on the top of my head. The other way to do this (and the way I personally prefer) is to draw a logic tree...


the first branch is the first pick, with 2/8 chance of picking a b&w and 6/8 of picking a color. The 2nd branch is the 2nd pick with the first pick removed so on the B&W first pick side, chances of getting the 2nd b&w is now 1/7 and getting a color is 6/7. On the first pick color side there is a 2/7 of getting a B&W and 5/7 chance of getting color.

The prob of each of the four outcomes (B&W, B&W) (B&W,Color) (Color, B&W) and (Color, Color) are figured by multiplying the probabilities down the tree as I did on the bottom. It is easy to see that all the outcomes except for (color, color) involve picking at least 1 B&W tv. Therefore all you need to do is add up the probabilities of the desireable outcomes (1/28 +6/28 +6/28 = 13/28) I think drawing the logic tree is quicker and easier to understand in a testing situation. If you have any questions about stuff like this, I know quite a bit of probability, although most people here are assholes and will flame you and call you lazy if you ask for hw help, which i never understood because its obvious im not just shooting in the dark when I try to get help. Good luck

to add, a shortcut to understand the xCy equation...


you can see that 8C2 cancels out to (8*7)/(2*1)

the secret here is there are 2 (the y value) numbers on top and 2 on the bottom. If you can understand the setup with another example without me showing the work but making a point...

9C5 = (9*8*7*6*5)/(5*4*3*2*1)

12C3 = (12*11*10)/(3*2*1)

starting to get it?

8C6 = (8*7*6*5*4*3)/(6*5*4*3*2*1) = 28 = 8C2 (theres the same amount of ways to leave 2 behind out of 8 as ways to take 2 away from 8 o_0)

Well explained, porsche944

wow...DS is being nice for once thanks p944, now that you've shown me like a diagram, its much easier...especially the way you treated 82C with the 2nd diagram...coz during gmat exam we dont have access to calc, so this helps.

next time i'll brush up on my math review lol.

Glad I could help

cool topic. i feel stupid.

I am very well acquainted with matters mathematical. I understand equations both the simple and the quadratical. About binomial theorems I'm teaming with a ton of news; with many cheerful facts about the square of the hypotenuse. I am very good at integral and differential calculus. I also know the scientific names of beings animalculous.

Looking at your replies I think I've discovered the simplest way to think about the problem. It's a sum of solutions over possibilities method. It works for a simple problem like this. The other typical method of summing individual event probabilities is perfectly valid, but perhaps too esoteric for a problem this tractable.

Counting the solutions goes as follows: Assume out of the 8 TVs you've already picked BW#1 TV, guaranteeing you satisfy the yes condition. There are 6 ways to choose a 2nd color TV, 6 solutions. This process is also true for BW#2 and the 6 color TVs, 6 more solutions. Add to the 6 and 6 the final unlikely case of picking both BW#1 and BW#2, 1 final solution. 6+6+1=13.

All that's left is to divide by the total number of combinations which would be 7+6+5+4+3+2+1=28.

13 solutions, 28 possibilities. 13/28.

but what if he was picking 12 tvs out of 350, and 47 of them are black and white? Memorising an oversimplified solution can lead to problems down the road

Memorizing a method isn't in itself problematic. Compulsively employing a single method amongst many with no regard for their relative applicability would be. However, all of us here are too smart and aware to fall into such an obvious trap. For those slightly behind the curve, my explicit statement that this method is more appropriate for such a tractable problem should readily imply to even the most dimwitted individual that the method is less appropriate for less tractable problems. It's painfully obvious that the method I used was arrived at by considering the problem's scale. The only static element was the way the problem was initially viewed which led to the method's selection.

But just to prove what an arrogant twat you are I'll proceed to apply the exact same method to your new problem with only some expanded handling of the greater dimension.

X 47 BWTV
Y 303 CTV
T 350 TV
Z 12 Selected

P(>= 1 BWTV) = (Positive Outcomes) / (Total Outcomes)

Assuming our selections contain exactly Q BWTVs, we wish to sum over all possibilities from Q=1 to Q=12. I'm avoiding for the moment the obvious shortcut that "at least 1 BWTV" provides and that is simply to count the number of times 0 BWTVs are selected and invert the result instead of counting all possible ways 1, 2, 3, etc BWTVs can happen. Finding the probability that exactly not 0 BWTVs are selected is equivalent to finding the probability that at least 1 BWTV is.

Assuming we select exactly Q BWTV(s), the remaining Z-Q selections must be CTVs. This means we have Y possible CTVs to choose from for the first team and Y-(Z-Q) CTVs to choose at the Z-Qth term. The terms would look like (Y)+(Y-1)+(Y-2)+.... but a regular, non-iterative formula is needed. The sum of a finite number of terms in a N+1 sequence is equivalent to the sum of the entire series minus the series beyond the limit. The total series sum is Y(Y+1)/2 and the series sum of the part beyond the limit is N(N+1)/2 where N = Y-(Z-Q). The sum of the terms is ( (Y)(Y+1) - (Y-Z+Q)(Y-Z+Q+1) )/2.

The sum of possibilities for each initial selection of Q BWTVs must then be multiplied by the number of ways one can arrive at Q BWTVs in each case. If Q=1 then there are X=47 ways to start, if Q=2 then there are X+(X-1) ways, Q=3 X+(X-1)(X-2) ways, and so on. The number is arrived at via the now hauntingly familiar N(N+1)/2 pattern; X(X+1)/2 - X-Q(X-Q+1)/2.

The sum of possible positive outcomes given exactly Q BWTVs selected is the multiple of the two terms:
( Y(Y+1) - (Y-Z+Q)(Y-Z+Q+1) )*( (X(X+1) - (X-Q)(X-Q+1) )/4.
( 303*304 - (291+Q)(292+Q) )*( (47*48 - (47-Q)(48-Q) )/4
(9140 - Q^2+538Q) * (95Q + Q^2)/4
Q( Q^3 - 633Q^2 + 41970Q + 868300 )/4

All that remains is to sum over Q from 1 to 12 (which I really don't want to do in closed algebraic form) and then divide by the total possible outcomes which is (T(T+1)-(T-Z)(T-Z+1))/2.

---

Shortcut: It's easier in this case just to find out how many possibilities zero BWTVs are selected and then invert the answer at the end. Selecting all CTVs would have Y(Y+1)-(Y-Z)(Y-Z+1)/2 possibilities out of the same total possibilities (T(T+1)-(T-Z)(T-Z+1))/2 as before.

(303*304-291*292) / (350*351-338*339) = 7140 / 8268. This is of course the probability of a negative outcome. Subtracting the negative result from unity we get: 8268/8268 - 7140/8268 = 1128 / 8268 or approximately 13.64%.

o_0

I was only implying that because the small scale of the problem allows him to bypass the equations, it doesnt mean he shouldnt know them...

but youre right, your ability to do math has proven that I am an arrogant twat?

....fukin nerds.... haha

Bypass the equations assumes the GMAT is interested in how well you memorized formulas in math class. These generalized exams seem to be more interested in thinking on your feet and methods that regular people can apply to slightly technical problems as opposed to giving you easy versions of problems to flex your N-dimensional general case equations.

I wouldn't go esoteric on this problem for the same reasons I wouldn't introduce the concept of linear independence to help a 6th grader solve a system of linear equations.

P.S. Just because I think that counting the cases of 0 BWTVs and inverting is yet simpler here goes. With 6 color TVs, picking 2 color TVs (and thus not at least 1 BWTV) would have 5+4+3+2+1=15 results out of 7+6+5+4+3+2+1=28 total. If there are 15 negative results then the other 28-15=13 results must be positive, thus 13/28 odds.

ok, so why am I an arrogant twat? And how was that proved by you solving a math problem? A math discussion would have been kinda fun, but you just randomly resorted to name calling for god knows why. But its funny because you've been on a tirade of 'something to prove" yet im the arrogant one, evidentally.

it is indeed easier to count the inverse in this case, but if you work with percentages you dont need to figure out the total possibilities. 1 - ((3/4)(5/7)) = 13/28. My whole reason for showing him the logic tree was that it is very flexable for "you choose ... out of ..." problems.

I would have done the large scale one the same way... 1-((303*302*301*300*299*298*297*296*295*294*293*292) / (350*349*348*347*346*345*344*343*342*341*340*339)). but thats not the point. The point is I think you are in fact an "arrogant twat" because you seem determined to prove that your methods are superior to mine while neither is true because they all point toward the correct answer. The only thing ive been trying to prove is your namecalling was out of the blue and uncalled for. I dont know anything about the GMAT specifically. I was just trying to help with the problem.

Um, no. Everyone was talking about math and having a civilized time of it until YOU came in and said "oh what if the problem is different. your idea is stupid. you are dumb." It was you that shit all over the thread and just because you don't like to be called on it doesn't change the fact that you were wrong.

Nobody knocked your method, they just added their own. No, but not you, you had to deride another method.

wow. actually it seems as though he gave a useful solution and was then complimented on his presentation by your own flesh and blood, as well as being graciously thanked by the original poster of the question. rising the issue of running into a different kind of problem with a greater level of complexity down the road wasn't an attack on the solutions that anyone had come up with, but merely an attempt to educate a total stranger who seemed receptive to any sort of suggestions that could be provided. after re-reading the posts here, it was kind of amusing that porsche mentioned that "most users are assholes and will flame you" and then he was subsequented flamed by you when you referred to him as an "arrogant twat". and the only reference you could possibly make to "deriding another method" would be when he said "Memorising an oversimplified solution can lead to problems down the road" which is probably the farthest thing from derision in this whole thread - in fact, id use the terminology "thoughtfully precautionary statement".

im not really sure where your deep animosity towards porsche comes from, and if this wasn't the internet and i were him, id demand an apology for such blatantly offensive and unwarranted attacks. so... i think i speak for everyone in this thread when i say stop being such a massive douche.

I think I speak for everyone in this thread when I say this should've been locked a page ago.

lol yes. i was happy with porsche's answer coz its what gmat is looking for...the equation he drew up. after that i stopped looking coz i didnt wanna overcomplicate my brain.

still, thanks though

thank you for this reply.

I've been around long enough to realize popularity has little to with correctness. Mob mentality all you like, but I stand by my statements. Just for fun let's actually go through the history of it and annotate.

#21-Frederf: I have observed the various methods presented and would like to add to the discussion because that's what gives me personal satisfaction. I see the sum of event probabilities method has been fleshed out to probably a greater extent than I would be able to do confidently and certainly adequate for this discussion. Since that method is well covered I will present an alternate method which is really just a reiteration and expansion on clarkma's method. I'll introduce my ideas with the preface that its value is in simplicity the other methods are correct and even much higher level to acknowledge their value and status.

#22-porsche944: I find fault with your method because it is overly simple and sacrifices general applicability despite that being the opposite of the goal of your method. I also incorrectly apply the fault of "memorizing a solution" which is not what you're doing at all, but I will apply this condescending accusation none the less, doubly-insulting as a falsehood. There is clearly not room in the world for a wide spectrum of methods as can be demonstrated by this problem I just made up. I also type in a lackluster way that is perceived as shouty, personal, and emotionally reactional.

$23-Frederf: I would like the cite the distinction between my explained method and the offensive accusation of "memorizing the answer." You have taken me to task for my ideas with a seemingly willful lack of understanding the motives behind my addition to the conversation. Not only is the notion that my idea is not welcome because it doesn't scale well and therefore some kind of mental crutch that would only serve to act as a landmine to the original poster when he comes across a similar problem, offensive, but it is also flat out wrong.

#25-porsche944: I state that even though both methods are valid it is inexcusable that the original poster be allowed to complete the problem without grasping my method's set of equations as opposed to your method's set of equations. This implicitly implies you were trying to do math "on the sly" with no regard for my version of rigorous mathematical treatment.

What's funny is I always get shit for trying to explain to people the rigorous or esoteric equations and theory behind generalized testing mathematics instead of the quick and dirty style that is so common. Then I hazard some generalized test math problem help that borders slightly on the side of Layman and I get shit for that too.

Honestly, just because you don't see why your words are offensive in academic discussion doesn't mean they aren't. When someone says, "Statement A with all the inherit limitations that Statement A implies" and you respond with "That's no good, what about the inherit limitations of Statement A?" it makes that person want to tear their hair out.

ahhh, it all makes sense now, I never realized you miss understood me. Here's how I intended post #22:

"While the solution here is enough to solve this particular problem, (heres where the confusion comes in) THE ORIGINAL POSTER SHOULD NOT ONLY MEMORISE THIS SOLUTION BECAUSE IT DOESNT ALLOW HIM TO SOLVE MORE DIFFICULT PROBLEMS.

I wasnt implying that you simply memorised the answer at all. At this point your retaliation to your misunderstanding confused the hell out of me. I never downplayed your ablities/methods only merely as bankie put it provided a caution toward the op.

I apoligize yadda yadda things got outta hand yadda yadda bygones being bygones and what not. You just have to understand that from my point of view and intention, you called me a twat out of the blue. And despite all that, debating math technique is just fun, because the open endedness of it never ceases to amaze me.

Hope THIS post was clear enough for you to understand

I think we all learned a lesson today - don't mess with ferderfaderf because he is the master of ALL DATAS! HHHHNNNNNNGGGGGG

Yeah, I knew what the criticism was. I didn't misunderstand it, I was annoyed you made it at all since it doesn't make sense. I still maintain that "counting all the positive outcomes and dividing by the total number of outcomes" is not some lameo one trick pony. It works for plenty of much harder problems.

Of course the particular way I counted doesn't scale well but the fact I did count isn't some landmine of a method. The counting method has the benefit of being easy to wrap one's head around and is a pretty nice fallback common sense approach for a high stress test. It's slightly less elegant and might take longer in some cases. Once the counting method is used, I thought it's pretty well understood that however you manage to count is up to the person, problem, and situation and of little consequence to the integrity of the method. My way of counting was intentionally reduced to as few words, steps, and thoughts as possible to make it seem straightforward with no regard to general applicability.

I think the ability to look at a problem and pick a technique that works well for it even if a slightly more complex version of the same problem would make that technique a pain is a good math skill. I mean at some point while solving a particular problem the solving process is going to diverge from the general set. I figure most people are going to be confused and frustrated by a generalized treatment and that absolute simplicity and confidence are better. Teaching the GF math problems has probably skewed my methods a lot.

In an interesting side note I found myself struggling to understand how many possibilities there were, should I multiply but picking the two things in the reverse? I was confused until I realized that the order of picking can't matter because it's perfectly valid to consider the picks simultaneous. We often use the concept of first selection, second selection to make things easier to think about but when I simply counted the totals it highlighted that there is no such thing as first and second.