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anonymous

  • 5 years ago

A student guesses the answers to 6 questions on a true-false quiz. Find the probability that the indicated number of guesses are correct: no more than 2(hint:no more than 2 means exactly 0 means exactly 1 or exactly 2)

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  1. anonymous
    • 5 years ago
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    put X = number he gets right. the probability he gets any one right is the probability he gets one any one wrong is \[\frac{1}{2}\] to get six wrong in a row multiply \[(\frac{1}{2})^6\] to get one right and 5 wrong it is \[\dbinom{6}{1}(\frac{1}{2})^6\] \[\dbinom{6}{1}\] is the number of ways to choose on item from a set of 6, so it is 6

  2. anonymous
    • 5 years ago
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    Thank you!!

  3. anonymous
    • 5 years ago
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    \[(\frac{1}{2})^6=\frac{1}{64}\] \[6\times \frac{1}{64}=\frac{6}{64}\] \[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\] \[1-\frac{7}{64}=\frac{57}{64}\]

  4. anonymous
    • 5 years ago
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    oops ignore the last line. just \[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\]

  5. anonymous
    • 5 years ago
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    man i am off. you also have to compute the probability that he gets two right. forgot about that one.

  6. anonymous
    • 5 years ago
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    it is \[\dbinom{6}{2}\frac{1}{64}\] and \[\dbinom{6}{2}=15\]

  7. anonymous
    • 5 years ago
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    so the correct answer is \[\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}\]

  8. anonymous
    • 5 years ago
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    sorry my mistake. not reading carefully.

  9. anonymous
    • 5 years ago
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    its okay! thank you! i got the answer correct :)

  10. anonymous
    • 5 years ago
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    great!

  11. anonymous
    • 5 years ago
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    are you good with pascal's triangle and other probabilities?

  12. anonymous
    • 5 years ago
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    sure got a question?

  13. anonymous
    • 5 years ago
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    lets see...what is the probability of getting 3 red lights at a traffic stop, when looking at five lights. assuming the common changes are red and green

  14. anonymous
    • 5 years ago
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    assuming these probabilities are each \[{1}{2}\ for each light and assuming they are independent?

  15. anonymous
    • 5 years ago
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    yeahh

  16. anonymous
    • 5 years ago
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    a rather broad assumption. so this is like saying if you flip a coin 5 times what is the probability you get 3 heads and one tail, yes?

  17. anonymous
    • 5 years ago
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    oh yeah, you could actually put it that way!

  18. anonymous
    • 5 years ago
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    then it is just \[\dbinom{5}{3}(\frac{1}{2})^5\]

  19. anonymous
    • 5 years ago
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    \[(\frac{1}{2})^5=\frac{1}{32}\] do you know how to compute \[\dbinom{5}{3}\]?

  20. anonymous
    • 5 years ago
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    no? haha i am not very great with math

  21. anonymous
    • 5 years ago
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    do you know how to write the first few levels of pascal's triangle?

  22. anonymous
    • 5 years ago
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    yeah i know how to do that

  23. anonymous
    • 5 years ago
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    1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

  24. anonymous
    • 5 years ago
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    there it is down to level 5 to compute each entry start at 1, then add the top two adjacent numbers.

  25. anonymous
    • 5 years ago
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    so to get the next level i would start at 1, then say 1+5=6 5+10=15 10+5=15 5+1=6 and so the next level would be 1 6 15 15 6 1

  26. anonymous
    • 5 years ago
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    okay...got it!

  27. anonymous
    • 5 years ago
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    but let us just say without writing the triangle i wanted to compute \[\dbinom{7}{3}\] the number of ways to choose 3 from 7. i would make a fraction and put in the top \[7\times 6\times 5\] in the bottom \[3\times 2\]tp get \[\frac{7\times 6 \times 5}{3\times 2}\] this has to be a whole number so cancel before multiplying.

  28. anonymous
    • 5 years ago
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    \[\frac{7\times 6 \times 5}{3\times 2}=7\times 5=35\]

  29. anonymous
    • 5 years ago
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    \[\dbinom{10}{4}=\frac{10\times 9 \times 8 \times 7}{4\times 3\times 2}\]

  30. anonymous
    • 5 years ago
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    again a whole number so cancel first and multiply last.

  31. anonymous
    • 5 years ago
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    \[\dbinom{10}{4}=10\times 3\times 7=210\]

  32. anonymous
    • 5 years ago
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    oh wow, okay i get it a little better now

  33. anonymous
    • 5 years ago
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    http://ptri1.tripod.com/

  34. anonymous
    • 5 years ago
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    more than you want to know

  35. anonymous
    • 5 years ago
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    okay i have another question that im having a hard time with

  36. anonymous
    • 5 years ago
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    shoot

  37. anonymous
    • 5 years ago
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    Suppose that you ask three friends to go to the mall. each one has a 0.80 chance of saying yes. What is the probability that all three friends will say they can go to the mall with you?

  38. anonymous
    • 5 years ago
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    ok that one is easy. \[(.80)^3\]

  39. anonymous
    • 5 years ago
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    assuming independence multiply the probabilities

  40. anonymous
    • 5 years ago
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    oh so its 0.512

  41. anonymous
    • 5 years ago
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    cool!

  42. anonymous
    • 5 years ago
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    you got it.

  43. anonymous
    • 5 years ago
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    wow thank you! are you good with the binomial theorem? haha

  44. anonymous
    • 5 years ago
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    or with the Fibonacci Sequence?

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