<p>THe probability that only two people in a room of three people are male is ?
It is from a practice test from the Gruber`s study guide</p>
<p>Let the three people be numbered 1, 2, and 3. First, calculate the total number of arrangements - each can be male or female, so 2x2x2=8 possible combinations. Now calculate the number of ways that exactly two can be male - this will be in the form of a combination since the order doesn’t matter. 3x2/2=3. So the probability is 3/8.</p>
<p>OK ,thank you.At first,i tought ‘‘what an easy question’’ .THen i realized i cant solve it 
BTW ,why do you devide 3x2/2 ?And ,in general,wy do we divede by 2 in such cases ?</p>
<p>Well if we just did 3x2, we would come up with six. Those possibilities would be:</p>
<p>1 and 2
1 and 3
2 and 1
2 and 3
3 and 1
3 and 2</p>
<p>Looks good, right? Actually, we listed each possibility twice - 2 and 1 is the same as 1 and 2, just in different order. So if order mattered, we wouldn’t divide by 2, but since order doesn’t matter in the instance, we do divide by 2 to get rid of the double answers. What we’re really doing is dividing by the number of ways two things can be arranged (2x1=2).</p>
<p>After dividing by two, we’re left with the three possibilities:</p>
<p>1 and 2
2 and 3
1 and 3</p>
<p>But when we devide by 3 or 4 or so long ?</p>
<p>If we were ordering things in sets of three, we would divide by the number of ways three things could be arranged (3x2x1=6) and so on.</p>
<p>It’s said that the order doesn’t matter so why it is 3/8? I think that it should be 1/4 in case that the order doesn’t matter - there is only one case in which there are 2 men and 1 woman. If the order DOES matter - I’m agreed but I don’t thing that the correct answer is 3/8 !</p>
<p>F F F
M F F
M M F
M M M</p>
<p>Well,the order doesnt matter ,and the answer is still 3/8 .</p>
<p>This problem does not require the combinatoric methods being used here. (In fact, NO sat problem ever does.)</p>
<p>It is equivalent to asking: if you flip a coin three times, what is the probablility of getting heads exactly twice. And it takes just a few seconds to list the possible outcomes:</p>
<p>HHH
HHT
HTH
HTT
THH
THT
TTH
TTT</p>
<p>I count 8. And 3 of the 8 have 2 H’s.</p>
<p>This method may seem simplistic, but I don’t get caught up in discussing whether order matters or not.</p>
<p>just think of it this way, list all the total possibilities of the people in a table, then circle all the ones that have ONLY TWO are male, then divide them</p>
<p>Boyan - when we say order doesn’t matter, that doesn’t mean it doesn’t matter which two of the three people are males, it just means it doesn’t matter how they are arranged in the room or the order they are lined up in. But having persons A and B, A and C, or B and C be the two males are still three distinct possibilities.</p>
<p>And of course one can solve it by just writing them out, but if you’re familiar with permutations/combinations it can be faster that way, and having two methods gives you a good way to check your work.</p>
<p>Good luck of listing all the possibilities if you were asked what is the probability that in a room of 50 people,48 are female :D:D:D</p>
<p>Most of the time the SAT questions on permutations/combinations/probability allow for writing out all possible arrangements.
If the numbers are too big (as in CR7_ManUtd’s example), you can always “model” the question with the smaller ones to refresh your memory as to what to do with them.
Examples.
Small numbers:
Questions from the BB - actors and roles; plumbers; cards (24 arrangements are invoved; still doable);
From the recent tests - number of games among a few teams; number of lines passing through a few points on two parallel lines.</p>
<p>Large numbers - only one question from the recent test comes to mind at the moment:
Number of lines passing through a bunch of points on a plane (the answer would be the same in 3D).</p>
<p>@gcf101 ,
how do you know about the problems from the recent tests :D</p>
<p>The SAT does not just suddenly start covering math at a higher level without announcing it. And I have been following the SAT for so long it is kind of depressing. In that time, I have never seen a problem that requires understanding of combinations vs permutations, etc… So I say list the possibilites and be done with it.</p>
<p>(But you DO need to know the counting principle: if there are X ways to do this and Y ways to do that, then there are X times Y ways to do this and that…)</p>
<p>It is just like the sequences…you can easily solve the problems if you know some formulas,but you can solve the even if you dont know any formulas.Just logical reasoning</p>
<p>the original question is kind of weird. you have to assume a probability distribution from which to pick males and females. it’s assumed to be 1/2 male, 1/2 female. but a binary choice does not imply 50/50 chance in general.</p>
<p>anyway, although pckeller is right about not needing to know combinations vs. permutations, it might make it easier. </p>
<p>given the combo male-male-female, you have 1/2 chance for picking each of those, so 1/2^3=1/8 that you picked that particular combination. There are three ways of ordering it, though, if you just move the F around: FMM, MFM, MMF. So it’s 3*1/8, where the 1/8 is from before. </p>
<p>you need to order it because you are treating the problem as if you are picking people (half of whom are male, the other half female, by assumption) from a large distribution and placing them in a room, and seeing how often theoretically you would pick 2 males, 1 female. </p>
<p>Using a permutation formula, ordering 3 elements, 2 of which are “the same” is 3!/(2!*1!)=3. See post #25 for explanation.
<a href=“http://talk.collegeconfidential.com/sat-preparation/657337-math-problem-2.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/657337-math-problem-2.html</a></p>
<p>If you know formulas and like concise answers, the “mathy” answer is just 1/2^3*3!/2!=3/8.</p>
<p>Even more (seemingly) complex questions can be answered by listing out all possible sets/arrangements:
<a href=“Tricky Math II Questions - SAT Subject Test Preparation - College Confidential Forums”>Tricky Math II Questions - SAT Subject Test Preparation - College Confidential Forums;
<p>Here’s another example:
Probability of getting a local call is 1/3;
probability of getting a long distance call is 2/3.
What is a probability of getting 2 different calls in a row?</p>
<p>It’s a hard one - even for the SAT II Math 2. Still, “listable”:
<a href=“http://talk.collegeconfidential.com/sat-subject-tests-preparation/611741-long-distance-math-level-2-a.html[/url]”>http://talk.collegeconfidential.com/sat-subject-tests-preparation/611741-long-distance-math-level-2-a.html</a></p>
<p>you could make a list there, but you either get local-long or long-local, so 2<em>1/3</em>2/3. i don’t know if i would call that hard. to me it seems like a good idea for someone who is making a list out of that to re-assess what the problem is asking.</p>