Quantcast XP Math - Forums - View Single Post - can someone answer these problems about probability and statistics?
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Old 07-28-2007   #5
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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=362. 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= 10233.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.