Pleaze help me with those two freak math

<p>|x|<em>|</em>|X|</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>this is something that i don't understand.</p>

<p>and this one</p>

<p>11 There are 6 red, 6 brown, 6 yellow, and 6 gray scarves packaged in 24 identical, unmarked boxes, 1 scarf per box. What is the least number of boxes that must be selected in order to be sure that among the boxes selected 3 or more contain scarves of the same color?</p>

<p>\ this is from the official college board\ so may be you will get the same type of qs. thanks for helping</p>

<p>The first one:</p>

<p>I think this is the answer although I have some doubts:</p>

<p>If there are 4 employees and 4 offices then there are a total of 24
(4 x 3 x 2 x 1) arrangements for the 4 emplyees to be assigned.</p>

<p>Karen can be in the offices with Xs in 2 of the 24 arrangements.
The same goes for Tina.</p>

<p>So te probability that they both get one is: 4/24 = 1/6.</p>

<p>(double check that because I'm not sure :$</p>

<p>Could also be 12 :/ (4P2 = 4x3=12) The number of the question would help.
I'd go for what mr. 17 got on #2. (im doing thisfrom observing the problem for a few seconds, and not as if i was solving it, so i probably got it wrong. :/</p>

<p>Q1. Here's a different way of looking at this problem.</p>

<pre><code>Think of Karen and Tina as 'red' employees, and the two 'X' offices as 'red' offices. We can allocate the offices in the order RedOffice1, RedOffice2, OtherOffice1, OtherOffice2 without affecting the answer to the problem.
</code></pre>

<p>For RedOffice1, probability of getting a 'red' employee = (# red employees) / (total #employees) = 2/4 = 1/2</p>

<p>If this happens, then
for RedOffice2, probability of getting a 'red' remployee =
(# unassigned 'red' employees) / (total # unassigned employees) = 1/3</p>

<p>Therefore, prob( 'red' employee in RedOffice1 AND 'red' employee in RedOffice2)
= (2/4)(1/3) = 1/6 , same answer as Stuck-on-1700 got in post#2 above.</p>

<p>Q2. There are four distinct colors. You can pick 2 each of the four distinct colors for a total of 8 boxes, without a 3<em>peat of any color. Pick a 9th box, though, and it <em>has</em> to be one of those four colors, each of which has already has 2 boxes picked - and you will have a 3</em>peat of that color.</p>

<p>Answer is 9 boxes.</p>

<p>For Number 2 I believe the answer is 12. Think of it this way. You are trying to select the LEAST number of boxes, so if you want to select the least number of boxes to get 1 of each color scarf, you will have to select 4 scarves(1 from each color). Then in order to select the least number of boxes for THREE of each color scarf it would be 3 x 4=12.</p>

<p>I hope this explanation helps.</p>

<p>0o0.Right it's asking you 'what is the PRoBABILITY...' probability is a fraction. Damn, I GOTTA read these problems more carefully! Optimizerdad has won the bout. I declare defeat on #1
Now i declare defeat on #2. It asks you how many boxes, not probability of the scarfs go into booxes-whatever. I HATE THESE WORD PROBLEMS!</p>

<p>Lol..calm down mozilla..</p>

<p>I'd calm down if I get a 700 in Math Sunday</p>