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)

- anonymous

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- 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

Thank you!!

- 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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## More answers

- anonymous

oops ignore the last line. just \[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\]

- anonymous

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

- anonymous

it is \[\dbinom{6}{2}\frac{1}{64}\] and \[\dbinom{6}{2}=15\]

- anonymous

so the correct answer is \[\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}\]

- anonymous

sorry my mistake. not reading carefully.

- anonymous

its okay! thank you! i got the answer correct :)

- anonymous

great!

- anonymous

are you good with pascal's triangle and other probabilities?

- anonymous

sure got a question?

- 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

assuming these probabilities are each \[{1}{2}\ for each light and assuming they are independent?

- anonymous

yeahh

- 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

oh yeah, you could actually put it that way!

- anonymous

then it is just \[\dbinom{5}{3}(\frac{1}{2})^5\]

- anonymous

\[(\frac{1}{2})^5=\frac{1}{32}\] do you know how to compute \[\dbinom{5}{3}\]?

- anonymous

no? haha i am not very great with math

- anonymous

do you know how to write the first few levels of pascal's triangle?

- anonymous

yeah i know how to do that

- anonymous

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

- anonymous

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

- 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

okay...got it!

- 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

\[\frac{7\times 6 \times 5}{3\times 2}=7\times 5=35\]

- anonymous

\[\dbinom{10}{4}=\frac{10\times 9 \times 8 \times 7}{4\times 3\times 2}\]

- anonymous

again a whole number so cancel first and multiply last.

- anonymous

\[\dbinom{10}{4}=10\times 3\times 7=210\]

- anonymous

oh wow, okay i get it a little better now

- anonymous

http://ptri1.tripod.com/

- anonymous

more than you want to know

- anonymous

okay i have another question that im having a hard time with

- anonymous

shoot

- 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

ok that one is easy.
\[(.80)^3\]

- anonymous

assuming independence multiply the probabilities

- anonymous

oh so its 0.512

- anonymous

cool!

- anonymous

you got it.

- anonymous

wow thank you! are you good with the binomial theorem? haha

- anonymous

or with the Fibonacci Sequence?

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