<p>oh dear this is a fight to the death lol, well im gonna insist back :P
i think the prob. is 2/9. think of it this way. i have two dice, and i throw them. i want to calculate the probability of getting a sum of eleven. either the first die shows a 5 and the other 6 or vice versa. so there is a 2 out of 36 prob. however with the phone calls, each one cant turn into the other on its own accord. they both are different and are not interchangeable. so local then int'l would be the same as int'l then local cuz each call can only be what it is. in this case u would only get a 2/9 probability. saying that the probabilty is 4/9 means this 2/9 * 2/9. this means that first time around either local then int'l then second time again either local or int'l. but thats not the case. you only do one of both and they're not related and they're different and you dont have a choice the second time u call (u need to call both types remember). if you did have a choice then it wouldn't be 1/3 * 2/3 * 2, cuz the first time u need to take the prob of both and then the second u need to consider only one so it doesnt become 4/9. once again i really think that its 2/9, 2/9, 2/9......</p>
<p>ok, so prove your version: build a sample space of 9, and set probability for a) 2 locals b) 2 long-distance c) different and get full probability of 1 in a sum</p>
<p>in my version from previous post
GU b
UG b
MU c
UM c
GM c
MG c
UU b
MM a
GG b</p>
<p>so probablity of two locals (a) is 1/9
two long-distance (b) 4/9
different (c) 4/9</p>
<p>1/9+4/9+4/9 =1</p>
<p>proved</p>
<p>now your turn :)</p>
<p>O...M....G guys, it doesn't even matter!
I now remember that I put 4/9, but the curve is -7 = 800. Who cares if you miss a question or two?</p>
<p>=====================</p>
<p>ok, so prove your version: build a sample space of 9, and set probability for a) 2 locals b) 2 long-distance c) different and get full probability of 1 in a sum</p>
<p>in my version from previous post
GU b
UG b
MU c
UM c
GM c
MG c
UU b
MM a
GG b</p>
<p>so probablity of two locals (a) is 1/9
two long-distance (b) 4/9
different (c) 4/9</p>
<p>1/9+4/9+4/9 =1</p>
<p>proved</p>
<h1>now your turn </h1>
<p>it doesnt work like that. look with the dice when u do a sample space for the first throw either a 5 or a 6 can out then in the second the one that didnt come out before should now (this is ofcourse assuming that we want a total of 11) but with the calls. the first time its either local or int'l, but the second time its only the one that didnt happen. now i dont think you need to draw a sample space cause your not really trying to find out how many different ways can you make the call so that you can actually calculate the probabilty of both. all you need to do is find out the chances of doing both. again order doesnt matter in this case as long as both happen its fine. so its not 4/9 its 2/9/ btw ur sample space is unusual you got three events going there G, M, and U when in fact its only two events... maybe thats why ur getting 4/9</p>
<p>It's 4/9, ask moodrets. lol.</p>
<p>OK. By asking for the probability that one is local and one is long distance, it's REAAAAAALLY asking - what is the probability that it's not "both local" or "both long distance." We can understand this better because we're all getting confused about order not mattering, etc etc etc. - Chances that it's both local = 2/3<em>2/3=4/9. Chances that it's both long distance = 1/3</em>1/3=1/9. Add those up = 5/9. Chances that it's NOT either of those = 9/9-5/9=4/9</p>
<p>====================</p>
<h1>OK. By asking for the probability that one is local and one is long distance, it's REAAAAAALLY asking - what is the probability that it's not "both local" or "both long distance." We can understand this better because we're all getting confused about order not mattering, etc etc etc. - Chances that it's both local = 2/3<em>2/3=4/9. Chances that it's both long distance = 1/3</em>1/3=1/9. Add those up = 5/9. Chances that it's NOT either of those = 9/9-5/9=4/9</h1>
<p>your calculations are incorrect. although both events are independent if u are gonna say same event two times then the first prob. different for the prob. of the same event second time around. so adding them like that is wrong.
back to sonicexx, your proof is hardly proof. you merely listed all possible 9 events (according to ur "3 event" samle space) and then u added them up well duh ur gonna get 9/9. but there are only two events not three. </p>
<p>also please remember that with dice, even if ur gonna throw two of them and they're outcomes are independent ; still both of the dice are the same. however u dont know how the probs. of both local and int'l calls are related cuz ur not given how each was calculated.</p>
<p>Dude, chillax, it's 4/9. INVENIAMVIAM did it the way I did, subtracting the probability you DON'T want from 1.</p>
<p>Um, the calculations are not incorrect - if someone calls the first time, 2/3 chance local. After you hang up, you get another call. Hmmmm, does that first call change the outcome of the second? No. Still 2/3 chance. Same way with long distance. And I didn't even do the subtracting method on the thing, I added 2/3<em>1/3 +1/3</em>2/3=4/9. Not that hard.
And everyone, stop relating the phone calls to dice and to whatever else - it's a phone call.</p>
<p>teehee the mathematical snobs collide....only on CC ;)</p>
<p>^ LOL. That made my night.</p>
<p>Unfortunately I can be considered one of them :(</p>
<p>I never thought ((2/3)<em>(1/3))</em>2 could ever produce such a heated discussion</p>
<p>i think we should all take a chill pill and watch tv or something lol</p>
<p>how about the simpsons?</p>
<p>or relevant to the discussion... NUMB3RS, anyone?</p>
<p>^ That's a badass show.</p>
<p>4/9ers represent!</p>
<p>
[quote]
I feel so good (I feel good, I feel good, I feel good)
Finally we understood
Boney M
[/quote]
</p>
<p>Baa-aack in post #58
<a href="local,local">quote=gcf101</a> + (ld,ld) + (local,ld) + (ld,local) = 1
one local + one ld = (local,ld) + (ld,local) = 4/9<a href="2/3">/quote</a>(2/3) + (1/3)(1/3) + <a href="2/3">b</a>(1/3) + (1/3)(2/3)** = 1</p>
<p>Two paths from here.
hookem's way: 1 - ((2/3)(2/3) + (1/3)(1/3)) = 4/9
IENVIMAN'S :D: (2/3)(1/3) + (1/3)(2/3) = 4/9</p>
<p>==================
Let's paraphrase this question to get around that solemn oath we were forced on us by the CB lest we bring trouble upon our beloved CC.</p>
<p>Using a coin as a classic model.
What's the P of getting a Tail-Head (in that order) after two throws?
(1/2)(1/2)
P of getting a Head-Tail?
(1/2)(1/2)
P of getting a Tail-Head or a Head-Tail after two throws?
(1/2)(1/2) + (1/2)(1/2)
..............
Let's now imagine that our coin is not fair (biased).
As I like saying in jest, same difference. :)
Assume P(Tail) = 2/5,
P(Head) = 3/5;
P(Tail-Head or Head-Tail) = (2/5)(3/5) + (3/5)(2/5).</p>
<p>What's VERY important here is that
P(Tail) + P(Head) = 1.
It's either Tail or Head (epic simile ;)).</p>
<p>============
sonicexx and int'lstudent'12 - it would be so much fun to see you lead competing debate teams into the battle!</p>
<p>funny you should say that, im president of my school's debate team :P and im effing persistent lol. but you guys are gonna have to ban me to end this discussion (or at least until the scores come out) lol plz dont do that. the thing is, i think "both" points of view (not the answers as you will see later on in this post) are correct. philosophically speaking (i know this is a math subject test!) the probability depends on how you look at it. if i said there was 1/3 chance that harvard called you saying u've been accpeted (fancy that!) and there was a 2/3 chance that yale called you (even better!), would you really care who called you first? multiplication has the property that says a<em>b = b</em>a, so when we say 1/3<em>2/3 -graphically speaking- (lol) its just the same as 2/3</em>1/3 (and mathematically too!) so pretty much 2/9 applies to either local then int'l or int'l then local. i dont think it matters which occurs first. because each event is only gonna happen once and they are not combined in any way, you should only be interested in their happening together which is <strong>holds breath</strong> 2/9 (squeaky voice).. but seeing as it took more than just math to solve, i hardly think this is the answer CB is looking for lol, but im gonna cross my fingers tho! now there's one more thing. (assume P local is 1/3 and P int'l is 2/3) there are two calls, the first one should be either local or int'l but not both. so the P for the first call is (1 - 1/3<em>2/9) thats 5/9. now when you come to calculate the P for the second call, you really cant becuase you simply dont have enough information! you dont know whether the first call was int'l or local. so now you have to assume two situations. either the first call was local then then second P is 2/3 or the first was int'l so the second P is 1/3. so your "overall" P is (1 - 1/3</em>2/3)<em>(2/3) + (1 - 1/3</em>2/3)*1/3 = 5/9 (1/3 + 2/3) = 5/9!!! so if you really care about order then its 5/9 not 4/9. but if you dont care about order i guarantee you its gonna be 2/9 ;)</p>
<p>i see your point now. You assumed that these are independent events. But they're not.</p>
<p>If call is local, probability is 1/3
If call is not local, AND THEREFORE is long-distance, probability is 2/3</p>
<p>so 1/3+2/3 =1</p>
<p>there are only these 2 probabilities in sample space (cause prob of all =1)</p>
<p>They haven't said that there was something like semi-local call, because then probability of our 2 would be less.</p>
<p>And you get correct answer now 4/9, but you divided by 2. Hey if you got 1 yellow and 1 blue bird, and question "when you're looking at 2 different birds in sequence, what's the prob that one will be yellow and one blue?", the obvious answer is 1.</p>
<p>But, thinking in your way, it could be either yellow, and then blue, or blue and then yellow. cause prob of each is 1/2, then </p>
<p>1/2+1/2=1, but order doesn't matter, so it's 1/2</p>
<p>it doesn't make sense.</p>
<p>and notice that, if you multiplied P(A^B)=P(A)<em>P(B)=1/2</em>1/2=1/4</p>
<p>it's even worse.</p>
<p>In this case correct is p=1, but we could get wrong answer if we are depending on pure numbers only.</p>
<p>i didnt assume that they were independent events, the question said so. besides the intersection of both events is indeed P(L)*P(I). so when you subtract the intersection from one, you get all chances of only one type of call. now you could argue that another type of phone call could have taken place, but you can't say this in these sort of situations. when we deal with probability in these questions we assume that the events that we are calculating are the only ones that can happen. come to think of it, there's a chance that the caller gets a heart attack just before making his call, or that earth gets hit by a meteor preventing any further discussion of the solution to this question :P
in addition you bird analogy is illogical, if there was only one bird of each type then the P of both is automatically set at 1. with the phone calls, though if you're gonna include 'extra information' then I assure you the probability is not going to be any of the 5 answer choices. also keep in mind that the question said that the only two events that could take place are local and international. adding the two probabilites gives you one which is a very aggravating observation. the phone calls can be considered independent - as stated by the question - if they simply ,as the name suggests, dont depend on each other. one could say that both calls belong to the same set of calls, in that case 1 out of every calls is local and 2 out of every three is int'l. therefore it is possible to say that the calls are independent simply because its up to fate (lol) or the callers discretion. or in more mathematical terms, you can't take the call out of the set thereby changing the P of the next call. so either the prob is 5/9 or 2/9</p>
<p>How about we get the actual question straight first.
Are you sure they were mutually exclusive or what?</p>
<p>The question stated the two events were independent.</p>
<p>I, of course, skipped this one and moved on to #50. :)</p>