<p>Are any of you good at solving hard probability problems? Did you struggle through any of it at all? I'm having trouble solving some probability problems (mostly on Math IIC tests) and I just can't "get it" easily. Sometimes I'm just confused about when to multiply, use factorials, permutation, combination, etc.</p>
<p>This is a good introduction to probability. A lot of the times, you just have to think the problems through. Don't just try cramming a bunch of formulas. Also, try working with easier problems first before you move on to hard ones.</p>
<p>It's like learning how to ride a bike. Before you can do that, you have to learn to crawl, walk, and run first, and then use training wheels before you can get into the hard stuff. It goes with everything you learn.</p>
<p>Here's a problem to get you started? How many ways can you arrange the word "PEN"?</p>
<p>The answer is 6, because there are 3 possibilities for the first letter, 2 for the 2nd, and 1 for the third. And of course, you'd multiply, because for each first letter, there are 2 possible 2nds and 1 third. Also, tree diagrams can really, really help!</p>
<p>Thanks a lot for the link! Yep, I know that diagrams can really help but a lot of time is wasted by just mapping them out. There's always a quicker, but more complicated, procedure that gives you the correct answer. It's just that I can't figure it out sometimes</p>
<p>Does anyone know the way to solve it with the calculator?</p>
<p>I remember learning about it the past year but forgot most of my information. It has something to do with nCr or nPr and then you can multiply additional info.</p>
<p>For example:</p>
<p>7C5 x (1/4)^5 x (3/4)^2 --> I remember that formula but what exactly does that mean again? And how could i possibly use something like that for 6-sided dye's again? For example: Possibility of exactly getting 2 three's</p>
<p>Can someone help me with this problem?:</p>
<p>How many ways can six people be seated at a round table?</p>
<p>If it's not 720, whack me hard.</p>
<p>how many ways can 6 ppl be seated at a round table</p>
<p>answer = 120
making a ring is equivalent to starting out with one predetermined element
so</p>
<p>1x5x4x3x2x1= 120</p>
<p>i do not think you should use 6! = 720 for this problem</p>
<p>Sun surf you are right, but what is your method for solving the problem?</p>
<p>ok i'm not good at explaining this but i'll try</p>
<p>there are 6 people sitting around
let's label the first person as x
to form a ring (round table) , you start at point x and you end up back at point x right
if you substitute 1 for point x, you will have 1 x 5 x 4 x 3 x 2 x 1 = 120 ways</p>
<p>it's just like doing 6 factorial (6! = 6 x 5 x 4 x 3 x 2 x 1 =720) but you substitute the 6 at the beginning with 1</p>
<p>use nPr when order matters , use nCr when order doesnt matter</p>
<p>wait what would the nPr equation be?</p>
<p>You would use 6!, but you would take into account that if everyone got up and moved one seat to the right, you would still have the same arrangement.</p>
<p>So there are 6!, or 720 ways of arranging the people.</p>
<p>But again, if everyone got up and moved one to the right, you'd still have the same arrangement. And there are 6 possibilities. So, if the people were letters, then ABCDEF is the same as BCDEFA, CDEFAB, DEFABC, EFABCD, and FABCDE. There are 6 possibilities for each, so you'd divide 720/6 = 120, in order to take out all the repeats.</p>
<p>-You can look at the Calling All Math Geniuses Thread too.</p>
<p>Oh okay I see now. So for ABCDEF there are 6 possiblities that result in the same arrangement and for ABCDFE there are also 6 possiblities that result in the same arrangment. etc.. Is that right? Thanks for explaining! =]</p>
<p>how to quote?? i can bold.</p>
<p>Tsenguun:</p>
<p>The format is:</p>
<p>[qu2ote=PERSON WHO SAID QUOTE]quote here...
[/quote]
</p>
<p>Remove the number 2 above and you should have it</p>
<p>
</p>
<p>Testing quotation....</p>
<p>JenniferJones. Right. That's why you have to divide by 6, to get rid of all the extras.</p>