<p>So I have a relatively easy math test tomorrow (not a final) on probability. I took notes in my textbook as I read the one relevant chapter on probability. Then I sifted through my class notes and further added to the list of things I should know, but don't. Finally I progressed into a former test from last year. I was able to solve all the problems on the test except for one. </p>
<p>How many different 5 digit zip codes can be formed from the numbers {1, 2, 3...9} if the middle digit is prime and the 5 digit number is even? (there is no repetition of digits). </p>
<p>So far I was thinking I could use the conditional probability equation, where:
P(A|B) = n(A[intersection]B)/n(B) for the part like... given ha the number is even... then the middle digit is prime. What I guess the crux of the problem is not being able to grapple my head around how to choose the other choices, I mean obviously there are four prime numbers in the given set, but I'm not sure how that affects the other four digits.</p>
<p>I also have two other questions of possibly similar content? from the textbook, and not nearly as important as the question above (I seem to just have trouble in one similar area ^.^) </p>
<p>How many terms are represented by the expression n(n-1)(n-2)...(n-r+1)? What is the 6th term?
This question I have trouble interpretting what the actual question is asking. Is that just asking for (n-6)? It just seems too simple. If its that simple, then the question seems a bit irrelevant. If it isn't I am familiar with the binomial theorem. </p>
<p>How many six-letter word can be made from HUMANOID if (these two parts are like the first question... and i have the same problem here)
a) the first letter must be a vowel
b) the word must begin and end with a vowel.</p>
<p>Just pretend you didn't have the restrictions when you set it up. For example, if the question were "How many different 5 digit zip codes can be formed with the numbers 1-9?" you'd set it up like this: 9 X 9 X 9 X 9 X 9 (Because of the fundamental counting principle.) The problem, though, limits the middle number to five, because there are five prime numbers in the set. The last one has to be four, because there are four even numbers in the set. Multiply it, and you get:</p>
<p>It's about the same as the first one. Use the fundamental counting principle.</p>
<p>HUMANOID has eight letters (and thankfully, no repeat letters) This time, however, you CAN'T repeat letters, so as you go to each subsequent space, you have to decrease the amount you are multiplying by by one. (A factorial)</p>
<p>Without any restrictions, this means that the number of different words you can get using the letters in HUMANOID is:</p>
<p>8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 40, 320</p>
<p>a: Just limit the first term you multiply to the number of vowels in HUMANOID</p>
<p>4 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 20, 160</p>
<p>b: Limit it the same way you did the last one. However, you must get rid of your vowels "first" because otherwise you will simply be left with one randomly chosen letter that might not be a vowel. Confusing, but here's what it turns out as:</p>
<p>4 X 3 X 6 X 5 X 4 X 3 X 2 X 1 = 8, 640</p>
<p>I have difficulty explaining probability with words...so I hope that helped! Good luck on your test!</p>
<p>Disney, there is no repetition allowed in the first question. However, it's pretty easy with casework.</p>
<p>Set the middle digit and the last digit. Find out how many possibilities. Then multiply by four to get the total number of possibilities for a given prime in position 3 (there are four even digits between 1 and 9). Then, multiply again by four to get the total total number (there are four primes between 1 and 9).</p>
<p>thank you everyone disneyguy, d_leet, and Baelor. lol. My test was 12:38 PM and right now it is 1:40 AM lol. Thanks guys. I ended up asking someone in school. Apparently the first q we went over in class teh day I was 20 minutes late to class. Coincidentally an extremely similar question appeared on the test.</p>
<ol>
<li>9<em>9</em>4<em>9</em>4 b/c there are only 4 prime numbers (2,3,5,7) and there are only four numbers at the end that can make an even number (2,4,6,8). All other numbers can fit into the 1st, 2nd, and 4th places.</li>
</ol>