<p>Here's a probability question that really bothered me from the online course....I'm just terrible at probability so am seeking some advice on how to solve them..</p>
<p>**Figure is 4 Squares with 2 X's. </p>
<p>So First Squares = X
Second Square = No x
Third Square = No x
Fourth Square = X**</p>
<p>Terrible squares, I know, but you get the point.</p>
<p>The figure above represents four offices that will be assigned randomly to four employees, one employee per office. If Karen and Tina are two of the four employees, what is the probability that each will be assigned an office indicated with an X ?</p>
<p>(A) 1/16
(B) 1/12
(C) 1/6
(D) 1/4
(E) 1/2</p>
<p>...</p>
<p>I was thinking maybe 2/4 x 1/3 x 1/2 x 1/1: I don't know, I'm very confused.</p>
<ol>
<li><p>When you have a complicated problem, try to solve a simpler problem. What is the probability that Karen alone will be assigned an office indicated with an X?</p></li>
<li><p>Once Karen takes an office marked with an X, what is the probability that Tina will get one with an X?</p></li>
<li><p>What's the probability that both will happen?</p></li>
</ol>
<p>You don't have to be good at probability to solve this problem. You should think one step at a time, solving simpler problems first. Okay, so having Karen and Tina be assigned spots is confusing. What if we just have Tina disappear and just use Karen? Solve the problem with just one person. Then try to find a way to get Tina involved: she has to pick from one of the three remaining seats left. So, really, the problem doesn't require much knowledge of probability, just the ability to use mathematical reasoning to solve a problem.</p>
<p>I see what you mean: this really simplifies the problem but my skills with probability are at best lackluster. So here goes, tell me if I'm wrong:</p>
<p>The probability of Karen in an X would be 2/4
The probability of Tina in an X would be 1/3</p>
<p>Therefore, 1/2 x 1/3 (the probability of both happening) would be 1/6...</p>
<p>However, don't I need to multiply this by 2 to account for Karen being the other X and Tina in hers (Karen and Tina switching places?)</p>
<p>Any help/advice would be appreciated.</p>
<p>Thank you dchow08 on the elucidating perspective -- it really simplified the problem but my approach is still rather undeveloped.</p>
<p>Since Karen should be seated first, the probability of Karen receiving a seat with an X is 2/4: 2 of the Xs out of the 4 seats.</p>
<p>Tina is then seated in the X which is a 1/3 chance that she will get that seat...</p>
<p>If Karen landed in the other X, Tina would still have a 1/3 chance to select the other spot.</p>
<p>I believe the fact that Karen could land in either of the X's is negligible because the 2/4 accounts for both X's so even if Karen were to land on another X, Tina would have the same probability.</p>
<p>^ Although, my thinking is a bit confused at this part...otherwise I believe the answer would be 1/6: I think I solved it another way as well using the earlier method</p>
<p>2/4 (Karen) x 1/3 (Tina) x 2/2 (can go anywhere) x 1/1 (can go anywhere)</p>
<p>Is the answer 1/6? If so, can you give me a more definitive reason why there would be no other multiplication involved and perhaps a better solution. I really need to clear up my facts on what TO DO and what NOT TO DO with these probability problems. </p>
<p>For example, suppose the question asked what is the total number of combinations that could arise from seating the 4...I would be pretty stumped at that simple probability question. </p>
<p>I would assume it is 4 x 3 x 2 x 1 but I would not know to account for duplicates or even if there are any etc, which is the exact problem I'm having at the moment. Can you provide me a guideline when to account for duplicates or not to.</p>
<p>I'm not totally sure whether I'm right, but this is how I reached my answer:</p>
<p>Think of how many ways the two girls can be situated in the two offices with X's. Then, divide that over how many possible different permutations (hint, hint) of office-assigning to the four employees is possible.</p>
<p>I don't believe it's 1/6. And because dchow is making you learn it the hard way, I'll refrain from saying what I got. I'm not even sure if it's correct. :P</p>
<p>Okay, I did it the tedious and long way because I did not know how to approach the problem with the accounting for the permutations to remove the variations of Karen/Tina which was confusing to me...</p>
<p>My long way:</p>
<p>I drew out a branching tree of all the possibilities...</p>
<p>1
2
3
4</p>
<p>Each leading # led to 6 variations 1234,1243,1324,1342 etc... Since there were 4 #s, 6 x 4 = 24 possible ways to arrange everybody.</p>
<p>I put #1 = Karen and #2 = Tina. Since there were only two trees that started with either 1 or 2, I started with those. Within each tree, 1 branch of 4 variations was already taken up by 1 or 2: Branch 1 only had 2 branches available with #2 in the same positions. This meant that only 2 from each tree would be eligible....this meant 4 variations </p>
<p>4/24 = 1/6</p>
<p>Variations:</p>
<p>1xx2
1342
1432</p>
<p>2xx1
2341
2431</p>
<p>These are the only 4 variations possible...and since there are 24 variations total: Answer = 1/6. </p>
<p>However, can somebody please clarify when to account for the variations and when not to -- I do not want to draw out these lengthy and excessive tree plots in the real SAT. Thanks! A quick explanation of when to divide by permutations and whatnot would suffice. Thank you so much.</p>
<p>answer is 1/6.
Total number of ways to arrange employees = 24
4x3x2x1
Total ways to arange employees in a way so that karen and tina are in the front
2x2x1x1
the first 2 being that you can chose from either karen or tina, and the last being that you already chose one of them, so theres only tina/karen left.
The middle is self explanatory, if you alreay chose 2 , theres only 2 people left. 2x1.
4/24 = 1/6</p>
<p>On the first bvox, you can only choose between tina or karen, so thats 2. Second box, you can only choose between 2 people since one(who ever it is) has already been chosen on the first box and you can only choose 2 for the second box because one of the 3 people left are tina/karen, and they cant be in the middle. So the third box, we only have 2 people left, and only one of them can be in that box. The last box we have either karen ot tina, but we already chose one of them from the first box, so only 1 is left. Bam 2x2x1x1.</p>
<p>Karen's chance of getting an X is 1/2.
Once Karen takes an X, Tina's chance of getting an X is 1/3. So the chance that both get X's is
1/2 x 1/3 = 1/6
Done.</p>
<p>lolilaughed, I don't understand your question about multiplying and dividing to account for duplicates.</p>
<p>What I meant was that some problems are not distinguishable so AB would be the same as BA. This problem does not pertain to this since 1234 and 2341 and different. I'm a bit mixed up when you can tell when you need to divide by so and so permutation to account for the non-distinguishable combo.</p>
<p>For example, AB is same as BA so in all the combinations we have to eliminate CD DC and AB BA, AC and CA so it doesn't account for both AB and BA but just AB. </p>
<p>Hopefully, that makes it a bit more clear.</p>
<p>Your help is appreciated!</p>
<hr>
<p>Also I don't fully understand why we don't multiply by 2 for the problem above. Was I correct in saying Karen switching places with Tina is negligible? I can't really picture this in my mind, only through the use of branching trees can I solve it.</p>
<p>Any light on that problem would help as well</p>
<p>You always have to divide to account for duplicates. For example, take this question: How many distinguishable ways can you rearrange the word BELLY? Normally, there are 5 x 4 x 3 x 2 x 1 ways to rearrange the word with 5 letters, but there are 2 L's. Let the first L be called L1 and the second L be called L2. If you listed out all the possibilities (without dividing anything), you would have duplicates of everything. One half of them will have the first L be L1 and the other half will have the first L be L2. </p>
<p>Here's an easy, simpler example: SEA.
SEA
SAE
ESA
EAS
ASE
AES</p>
<p>Now, SEE. If you distinguished the first E, from the second E, you would get 6distinct possibilities.
S (E1) (E2)
S (E2) (E1)
(E1) S (E2)
(E1) (E2) S
(E2) S (E1)
(E2) (E1) S</p>
<p>But if you didn't distinguish them, you would get this
SEE
SEE
ESE
EES
ESE
EES
You would have duplicates, because the E's aren't distinguished.</p>
<p>^ I see!! Thanks for the tip; How do I tell what to divide by to account for the duplicates?</p>
<p>For example, SEE is clear to divide by 2 because there are two E's, but on another problem I don't believe it would be so obvious. I can't refer it to exactly at the moment, but I remember there was an SAT question of the Day similar to this that I didn't know if I had to divide to account for duplicates or not to. </p>
<p>Is the basic underlying rule to see whether it is distinguishable or not?</p>
<p>Your replies are very helpful and insightful! Thanks dchow.</p>
<p>You flip 5 coins. What's the probability of getting exactly 3 heads?</p>
<p>You know that there are 32 possibilities (2^5).</p>
<p>Now you need to find out how many ways you can get 3 heads and 2 tails.</p>
<p>For example,</p>
<p>HTHTH and HHTTH are two ways. How many ways are possible without repeating? That's pretty much the same as asking how many ways can you rearrange HHHTT distinctly?</p>
<p>If you don't distinguish the H's and the T's, HTHTH will repeat itself 12 times. The H's will repeat 6 times because HHH, if distinguished, can be arranged 6 times, and the T's can be arranged 2 times. With duplicates, there are 120 total ways: the first term could be one of the 3 H's or two of the 2 T's, then after that's taken the second term has only four options, then 3, then 2, then 1. 5x4x3x2x1 = 120.</p>
<p>Now since there are 12 duplicates for each of the distinct possibilities, there are only 120/12, or 10, distinct possibilities of 3 heads and 2 tails. Again, there are 32 possible outcomes, so the probability of getting exactly 3 heads and 2 tails is 10/32 = 5/16. Okay, that might be too difficult right now, but the point is that you have to divide to eliminate duplicates.</p>
<p>Thanks, I see that that method only works with a distinct set though...
What I mean is that it only works if it says 3 heads and 2 tails exactly.</p>
<p>What if it were to say exactly 3 heads (not specifying tails)? It would be HHHxx
or
What if it were to say at least 2 heads?</p>
<p>I remember I learned all this in pre-calculus but I long forgot...</p>
<p>What if it were to say exactly 3 heads (not specifying tails)? It would be HHHxx (A coin is either heads or tails.)
or
What if it were to say at least 2 heads? (you'd have to find the probability of 2 heads exactly, the prob. of 3 heads exactly, 4 heads exactly, and 5 heads, and add them all up).</p>