Can someone answer these problems about probability and statistics?

1. A truck driver can take three routes from city A to city B, four from city B to city C, and three from city C to city D. If when traveling from A to D, the driver must proceed from A to B to C to D, how many possible A to D routes are available?





2. If a multiple choice test consist of 5 questions each with 4 possible answers of which only one is correct. How may different ways can a student check off one answer to each question? How many ways can a student check off one answer to each question and get all the questions wrong?





3. How many three-digit numbers can be formed from the digit 0, 1, 2, 3, 4, 5, 6 if each digit can be used only once? (b) How many of these are odd numbers? (c) How many are greater than 330?





4. From a deck of cards, in how many ways can you pick the following card – A , K, Q, J





5. How many different ways can 4 red, 4 yellow, and 3 blue bulbs be arranged in a string of Christmas tree lights with 11 sockets?

Can someone answer these problems about probability and statistics?
This is basically the counting principle--you just multiply the individual probabilities to find the total.





#1: 3 x 4 x 3 = 36 possibilities





#2: there are three ways of getting each question wrong, so 3x3x3x3x3 is 243





#3: (a) there are seven possibilities for the first number (0 1 2 3 4 5 or 6), 6 possibilities for the second number (take off whatever was used for the first) and 5 for the third number. So 7x6x5 is 210





(b) Same process, but the last number has to be odd. so the last number is 3 instead of 5 7x6x3 so 126





(c) same process, but hte first two have to be bigger than three. So the first one can be a 4 a 5 or a 6, same with the second. Except the second can't match, so there are only two possibilities. The last number can't be 0, so there are six possibilities


3x2x6=36





4. Need more details. In that order, or just those 4 cards?





5. If you don't have to account for the actual color, just how many orders of bulbs, it is 11!


Otherwise, you would have to do permutations and I'd use a calculator (which I don't have here)
Reply:325.74
Reply:The best way to do these is to diagram them. You'll then be able to count the permutations.





By diagramming simple problems like these, you learn the probability formulas that enable you to solve more complicated ones.





1. Draw A, B, C, and D. Draw three different lines from A to B, four from B to C and three from C to D. For each of the three routes from A to B there are four routes from B to C. Therefore, there are 12 (3x4) ways to get from A to C. For each of these 12, there are 3 ways to get from C to D. Therefore, there are 36 (3x4x3) different ways to get from A to D.





Continue in the same manner with the others.
Reply:I just recently took AP Statistics so i might be able to help you out.





1. First you have 3 different routes, then 4 different routes and finally 3 different routes. By multiplying 3x4x3 you get all the possible routes. A-D=36





2. For each test question you have 4 different choices, so 4^5 or 4x4x4x4x4 would give all the possible ways of checking of the answer choices. 4^5=1024


To get all of the answers wrong, you would subtract the number of times you don't get them all wrong which is 1. So 1024-1= 1023





3.For this problem you would want draw three lines: _ _ _ this represents the three digits. Then there can be 7 different numbers for the first digit and only 6 and 5 for the next two be cause you can not repeat numbers. So to find the total amount of three digit numbers you would simply multiply. 7x6x5=210.


Out of a possible 210 times, the odd numbers will appear when either a 1, 3 or 5 is placed as the final digit. Since there is 7 different numbers the probability that any given number will be the final digit is 1/7 of the time, so 210x(1/7)=30. However there are three different odd numbers to 30x3=90. Hence forth, out of 210 numbers, 90 will be odd.


I'm not quite sure on (c).





4. Just as the last question you are choosing a certain number of specific items (this time cards). In this question you are choosing from Ace, King, Queen and Jack of Clubs, Diamonds, Hearts and Spades. This means that you are choosing 16 cards with out replacement (once you pick a card you can not pick it again since it is no longer in the deck). So, _ _ _ _ 16x15x14x13=43680 different ways of picking those cards.





5. Again a choosing items question, this time with 11 items (4+4+3). Simply put you have 11 spots with 11 lights, each light gets a spot. This is called 11! or 11 factorial which looks like: 11x10x9x8x7x6x5x4x3x2x1= 39916800.





Hopefully this was helpful if you have any other statistical questions, maybe i can help you out again.