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Thank you!!

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

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

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

sorry my mistake. not reading carefully.

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

great!

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

sure got a question?

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

yeahh

oh yeah, you could actually put it that way!

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

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

no? haha i am not very great with math

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

yeah i know how to do that

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

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

okay...got it!

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

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

again a whole number so cancel first and multiply last.

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

oh wow, okay i get it a little better now

http://ptri1.tripod.com/

more than you want to know

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

shoot

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

assuming independence multiply the probabilities

oh so its 0.512

cool!

you got it.

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

or with the Fibonacci Sequence?