<li><p>A team of 3 si picked at random from 5 ppl: A B C D E. What’s the probatility that both A and B will be on the team? (3/10) ( How to do that?)</p></li>
<li><p>Six card distribute to 3 ppl. Each at least gets one. How many different outcomes? ( 10)( Why)</p></li>
</ol>
<h1>1 you can systematically list to find the solution</h1>
<p>list: A, B, C, D, E</p>
<p>A can pair with 4 (including B, C, D or E), B with 3 (C, D or E), C, with 2 (D or E) or D with 1 (E). 10 pairs. </p>
<p>AB can be paired with any other third person... ABC, ABD, or ABE, for 3 combinations with A and B in them.</p>
<h1>2 I don't have any idea, except that it looks a little too hard to be an actual SAT problem.</h1>
<h1>2</h1>
<p>2 2 2
1 2 3
1 3 2
3 1 2
3 2 1
2 3 1
2 1 3
1 1 4
1 4 1
4 1 1</p>
<p>Each column represents a different person. But there has to be a simpler way to do it...</p>
<p>Where did you find these?</p>
<p>Are these real SAT questions? Where did you get them from? This stuff is scary:eek:</p>
<h1>1. An alternative way of solving this (and perhaps simpler) is</h1>
<p>Think of it as picking the two people who won't be on the team; the ones left behind <em>will</em> be on the team. Also, for simplicity, think of A &B as 'red', and C/D/E as 'blue'.</p>
<p>Prob( A & B on the team of 3) = Prob( neither A nor B is on the 'non<em>team' of 2) = Prob( only blues in the non</em>team)
The required probability is
Prob( 1st of 2 is blue) * Prob( 2nd of 2 is also blue)
= (3/5) * (2/4) {since after picking the first blue, you have a pool of four remaining candidates, two of whom are blue}
= 3/10 .</p>
<p>they are the real SAT problem published in past ten years. I got them from Testmasters.</p>
<p>"A can pair with 4 (including B, C, D or E), B with 3 (C, D or E), C, with 2 (D or E) or D with 1 (E). 10 pairs. "</p>
<p>Thnks. But i don't understand why is "pair"? Shouldn't it be "triple"? You explaination & answer are right.</p>
<p>"1. An alternative way of solving this (and perhaps simpler) is</p>
<p>Think of it as picking the two people who won't be on the team; the ones left behind <em>will</em> be on the team. Also, for simplicity, think of A &B as 'red', and C/D/E as 'blue'.</p>
<p>Prob( A & B on the team of 3) = Prob( neither A nor B is on the 'non<em>team' of 2) = Prob( only blues in the non</em>team)
The required probability is
Prob( 1st of 2 is blue) * Prob( 2nd of 2 is also blue)
= (3/5) * (2/4) {since after picking the first blue, you have a pool of four remaining candidates, two of whom are blue}
= 3/10 ."
Woo, it's awesome. thnks</p>
<p>for #2, the answer's explanation is 3+6+1=10
it devided to three catagories. Since it is "at least", so it can happen to be when ppl get 4 cards (3), 3 cards (6), 2 cards (1). But I still didn't get it. My Combination, Permutation and Prob are really bad. I didin't actually learn those. Are there some way can learn those?</p>
<p>""A can pair with 4 (including B, C, D or E), B with 3 (C, D or E), C, with 2 (D or E) or D with 1 (E). 10 pairs. "</p>
<p>Thnks. But i don't understand why is "pair"? Shouldn't it be "triple"? You explaination & answer are right."</p>
<p>Here it happened because "2 chosen from 5" and "3 chosen from 5" happen to yield the same answer, which is 10. </p>
<p>So, in this case: </p>
<p>How many TRIPLE COMBINATIONS are there?
1. ABC
2. ABD
3. ABE
4. ADE
5. ACD
6. ACE
7. ADE
8. BCD
9. BCE
10. CDE</p>
<p>How many TRIPLES are there that include AB
1, ABC
2. ABD
3. ABE</p>
<p>So, by comparing the two tables, we see that only 1,2,3 are common out of ten possibilities. Let's remember that probabilitie is number of positive outcome (success) over total possibilities. In this case, it is 3 over 10. </p>
<p>As far as learning probabilities for the SAT, you have a few choices: </p>
<p>1.You can learn how to LIST the combinations pretty fast. In general, ETS won't ask students to list a LARGE number od number. </p>
<ol>
<li><p>You can learn how to calculate combinations by using the formulae that use factorials. This would look like Combination(n,k) = n!/k!(n-k)! or in this case, Combi(5,3) is 5!/3!2! or 5<em>4</em>3<em>2</em>1 / 3<em>2</em>1 * 2<em>1 or 5</em>4/2<em>1 or 10. The same exercise for the Combi(3,2) would yield 3</em>2<em>1 / 2</em>1 * 1 or 6/2 or 3. </p></li>
<li><p>You can learn how to use your calculator. Every guide worth its salt will show you how to enter combinations and permutations. </p></li>
</ol>
<p>As far as what to enter in your calculator, there is where the reasoning comes into play. Always make sure to understand the question correctly. The next step is to apply the method you decided to learn. As usual, this is done by working through exercises until it becomes "easier". There are no shorcuts and magic explanations ... you just have to dig in with abandon.</p>
<p>OK, here it is a lot easier (to me).</p>
<ol>
<li><p>Denominator: C(5,3) (Ways to choose team of 3 from 5)
Numerator: You have three slots on the team. Two of these are reserved for A and B. One slot is left, with 3 people competing for it. Thus, there are 1<em>1</em>3 = 3 ways to choose a team w/ both A and B. thus, the desired prob. is 3/10</p></li>
<li><p>This is equivalent to finding the number of solutions to (1) a+b+c = 6, for a, b, c positive integers. Set a = x+1, b = y+1, c = z+1, with x, y, z nonnegative integers; then substitute into the original equation, yielding
(2) x+y+z = 3. The number of solutions to this in nonnegative integers to (2) is the same as the number of solutions in postive int's to (1) because of the def's of x, y, z. Now the problem trivializes; consider a group of three balls and two dividers as below:
<em>|</em>*|
We realize that the dividers partition the three balls into groupings which can be variable values( ex. for this arrangement, if we stipulate that from left to right the variables are x, y, z, we have x=1, y=2, z=0). Thus, the desired number is C(3+2, 3) = C (5,3) = 10</p></li>
</ol>
<p>(I explained a number of common techniques here for completeness; note the balls and sticks trick, as it arises commonly in count-the-solutions problems)</p>
<p>Ah, no wonder I kept on getting a huge number. I read it as six unique cards... horrible SAT question. This would certainly have been disqualified from any AMC competition.</p>
<p>for probability questions, just use logic. they are extremely simple if you forget about trying to be mathematical (that is, if you aren't too good with permutations, combinations, etc.) and just use simple logic to solve the questions. Probability questions on the SAT are never hard, and if you get stuck, you can always find all the cases in like 2 minutes.</p>