<p>probability must be 4/9, suppose we were expecting 2 calls from india (from grandma -G, and uncle -U) and one from our area (mum -M), and they could be repeated. we have options
GU
UG
MU *
UM *
GM *
MG *
UU
MM
GG</p>
<p>so we have 4/9. If order specified, we would have 2/9.</p>
<p>As I remember, there was question about limit, and there was parabola. Parabolas don't have limits, right?</p>
<p>the probability question has to be 2/9 because order simply does not matter. the question asked the probability of both events happening. when you multiply both probablities it is already possible their one can occur first or later. so it would be 2/9.</p>
<p>no, if there would be something like first call local, 2nd non-local, then we will have only 2 possibilities out of 9. Order doesn't matter, so we have 4 out of 9. Order matters=probability is less.</p>
<p>Correct 4/9</p>
<p>dont remember linearization. 2/9 seems like a joe bloggs answer though like CB expected you to just multiply em together..</p>
<p>i thought the answer was 2/9. cos for independent event P(AnB) = P(A) X P(B).</p>
<p>^ That's what I put as well -- @ sonic it was =25</p>
<p>Yeah, 2/9's does seem like a Joe Bloggs answers, esp. for a number 49. Dayum, too bad I failed for the Joe Bloggs answer.
When I was driving back home, I felt so bad when I went over the problems in my head and knew how to do them in the car but not in the testing room.</p>
<p>no, these weren't independent events (mathematicaly), independent are for example that joe will drink milk this morning and eva will drink milk this morning. In that situation, you can multiply, but not in this question.</p>
<p>Just try to workout sample space of 9 situations (as I did before), and check the ones question was asking. </p>
<p>IT IS 4/9!!!!!!!</p>
<p>the answer cannot be 4/9 order does not matter they simply want to know the probability of both events happening. its 2/9. college board didnt develop Joe Bloggs....</p>
<p>the question stated "independent events" iirc</p>
<p>sonice is correct - the answer is 4/9. The P(A+B) = P(AB) only works if the two events occur in sequence. Try drawing a tree diagram.</p>
<p>how much has the curve been in recent administrations? around -7 = 800?</p>
<p>that is 3rd time I'm saying </p>
<p>ORDER DOESN'T MATTER=PROBABILITY IS HIGHER</p>
<p>ORDER MATTERS=PROBABILITY IS LESS</p>
<p>just think.</p>
<p>probability that you'll throw 6 on a dice is 1/6. Probability, that you'll throw 2 is 1/6. So, probability that you'll throw 6 AND THEN 2 is 1/36. Probability that you'll throw 6 OR 2 must be 2 times as great.</p>
<p>i dont know where this is headed to lol, but i still think its 2/9. yes it is possible for one type of call to occur before another, but they r not interested in order. what you say about order meaning less probability is only true when you are asking for a specific scenario out of possible many. in this case however they said that there are two possible calls, and they said their respective probabilities were independent. all they asked was the chance of them happening. saying that the order matters suggests that one of the behind the scences working out would have been, "how many ways can the calls be made" but this is not the case. if the probabilty of it raining tomorrow was 1/3 and the day after 2/3 and the events were independent, then the probability of it raining on both days is simply 1/3 * 2/3 even though both cnt happen on the same day and that they are in sequence, at the end you are only interested in them happening.</p>
<p>intl
in your example if we will try to make classic probabilistic model</p>
<p>Z1={1,2,3} Z2={2,3,4} 1,2 - raining 3,4 -not raining. as you said, probability of raining in 1st day is 2/3, in second 1/3</p>
<p>possibilities:</p>
<p>day 1 :...1...2...3...2...3...1...2...3...1
day 2 :...2...3...4...2...3...4...4...2...3</p>
<p>sample space=9</p>
<p>probability
both raining= 2/9
raining only 1st day=4/9
raining only 2nd day=1/9
both dry= 2/9</p>
<p>OH NO WE WERE ALL WRONG!!!!!!!</p>
<p>probability of different events is 5/9!!!</p>
<p>DAMN</p>
<p>^ HAHAHA 5/9 was an answer choice, I think, but I have no understanding of your analogy whatsoever.</p>
<p>So, guys, I second blanche's question. What's the predicted curve? 43/50= 800? lower? higher?</p>
<p>WHOA, revelation! I think I remember how I did this problem and got 5/9.</p>
<p>I think you can use Venn diagrams....
It's whatever that rule is about adding stuff up and then subtracting the probability that you counted twice.....found it: SparkNotes:</a> SAT Subject Test: Math Level 2: Group Questions</p>
<p>wait, I'm thinking of a totally different problem lol</p>
<p>this is classic model, so every event is made to have equal probability of happening, i.e if you know the probability of one event is 75%, 2nd is 25%, then you made 4 digits/letters/whatever one to be 2nd event, and 3 to be second.</p>
<p>Oh, ok, gotcha. I still have no recollection of that problem, so I'm no help here haha</p>
<p>:( i bet the majority of people put 2/9</p>
<p>I KNOW NOW! </p>
<p>IT WAS 4/9, as I've written before :)</p>
<p>
[quote]
probability must be 4/9, suppose we were expecting 2 calls from india (from grandma -G, and uncle -U) and one from our area (mum -M), and they could be repeated. we have options
GU
UG
MU *
UM *
GM *
MG *
UU
MM
GG</p>
<p>so we have 4/9. If order specified, we would have 2/9.
[/quote]
</p>
<p>Example with rain is completely different, cause it puts probability for certain days.
5/9 would be right if our question #49 looked like "probability, that first call is local... "
But it wasn't :)))))))</p>
<p>So it's 4/9, 4/9, 4/9 :) <em>she smiles proudly</em></p>