anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
Thank you!!
anonymous
  • anonymous
\[(\frac{1}{2})^6=\frac{1}{64}\] \[6\times \frac{1}{64}=\frac{6}{64}\] \[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\] \[1-\frac{7}{64}=\frac{57}{64}\]

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anonymous
  • anonymous
oops ignore the last line. just \[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\]
anonymous
  • anonymous
man i am off. you also have to compute the probability that he gets two right. forgot about that one.
anonymous
  • anonymous
it is \[\dbinom{6}{2}\frac{1}{64}\] and \[\dbinom{6}{2}=15\]
anonymous
  • anonymous
so the correct answer is \[\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}\]
anonymous
  • anonymous
sorry my mistake. not reading carefully.
anonymous
  • anonymous
its okay! thank you! i got the answer correct :)
anonymous
  • anonymous
great!
anonymous
  • anonymous
are you good with pascal's triangle and other probabilities?
anonymous
  • anonymous
sure got a question?
anonymous
  • anonymous
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
anonymous
  • anonymous
assuming these probabilities are each \[{1}{2}\ for each light and assuming they are independent?
anonymous
  • anonymous
yeahh
anonymous
  • anonymous
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?
anonymous
  • anonymous
oh yeah, you could actually put it that way!
anonymous
  • anonymous
then it is just \[\dbinom{5}{3}(\frac{1}{2})^5\]
anonymous
  • anonymous
\[(\frac{1}{2})^5=\frac{1}{32}\] do you know how to compute \[\dbinom{5}{3}\]?
anonymous
  • anonymous
no? haha i am not very great with math
anonymous
  • anonymous
do you know how to write the first few levels of pascal's triangle?
anonymous
  • anonymous
yeah i know how to do that
anonymous
  • anonymous
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
anonymous
  • anonymous
there it is down to level 5 to compute each entry start at 1, then add the top two adjacent numbers.
anonymous
  • anonymous
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
anonymous
  • anonymous
okay...got it!
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\frac{7\times 6 \times 5}{3\times 2}=7\times 5=35\]
anonymous
  • anonymous
\[\dbinom{10}{4}=\frac{10\times 9 \times 8 \times 7}{4\times 3\times 2}\]
anonymous
  • anonymous
again a whole number so cancel first and multiply last.
anonymous
  • anonymous
\[\dbinom{10}{4}=10\times 3\times 7=210\]
anonymous
  • anonymous
oh wow, okay i get it a little better now
anonymous
  • anonymous
http://ptri1.tripod.com/
anonymous
  • anonymous
more than you want to know
anonymous
  • anonymous
okay i have another question that im having a hard time with
anonymous
  • anonymous
shoot
anonymous
  • anonymous
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?
anonymous
  • anonymous
ok that one is easy. \[(.80)^3\]
anonymous
  • anonymous
assuming independence multiply the probabilities
anonymous
  • anonymous
oh so its 0.512
anonymous
  • anonymous
cool!
anonymous
  • anonymous
you got it.
anonymous
  • anonymous
wow thank you! are you good with the binomial theorem? haha
anonymous
  • anonymous
or with the Fibonacci Sequence?

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