Question

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)

Answer 1

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

Answer 2

Thank you!!

Answer 3

\[(\frac{1}{2})^6=\frac{1}{64}\]
\[6\times \frac{1}{64}=\frac{6}{64}\]
\[\frac{1}{64}+\frac{6}{64}=\frac{7}{64}\]
\[1-\frac{7}{64}=\frac{57}{64}\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

sorry my mistake. not reading carefully.

Answer 9

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

Answer 10

great!

Answer 11

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

Answer 12

sure got a question?

Answer 13

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

Answer 14

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

Answer 15

yeahh

Answer 16

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?

Answer 17

oh yeah, you could actually put it that way!

Answer 18

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

Answer 19

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

Answer 20

no? haha i am not very great with math

Answer 21

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

Answer 22

yeah i know how to do that

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

okay...got it!

Answer 27

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.

Answer 28

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

Answer 29

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

Answer 30

again a whole number so cancel first and multiply last.

Answer 31

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

Answer 32

oh wow, okay i get it a little better now

Answer 33

http://ptri1.tripod.com/

Answer 34

more than you want to know

Answer 35

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

Answer 36

shoot

Answer 37

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?

Answer 38

ok that one is easy.

\[(.80)^3\]

Answer 39

assuming independence multiply the probabilities

Answer 40

oh so its 0.512

Answer 41

cool!

Answer 42

you got it.

Answer 43

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

Answer 44

or with the Fibonacci Sequence?