posted twice, sorry
I di #4 already, scroll down
1.) An octave contains twelve different musical notes (in Western music). How many different eight note melodies can be constructed from these twelve notes if:
(a) no note can be used more than once?
(b) any note can be used as often as you please?
Ans for a) a permutation
b) a combination
I believe that your answer for b
is the correct answer for a
These in a are 12 different notes, like 12 different people, who are arranged in groups of eight in any order.
This is the very definition of combinations of 12 different things taken eight at a time. So I say 495 for a
Now for b, we are less restricted. we can take any of the 12 eight times
THAT is in fact
which is very big indeed !
3. The chance that Ariel sees the movie Norbit is 45%. The chance that Brandon sees the movie is 50%. The chance that Ariel and Brandon both see the movie is 30%.
(a) If Brandon sees the movie, what is the chance that Ariel also sees it?
Answer: P(A&B)=(.45)(.50)= 22.5%
p(A) = .45
p(B) = .50
p (A and B) = .30
now we want p(A if B)which is written p(A/B)
Conditional probability definition:
p(A/B) = p(A and B)/p(B)
p(A/B) = .30/.5 = .60
5. Widgets are produced at a certain factory by each of three machines A, B and C. These machines produce 1000, 600 and 400 widgets per day, respectively. The probability that a given widget is defective is 4% for one produced by Machine A, 3% if produced by Machine B, and 2% if produced by Machine C. Suppose that the Widget Inspector selects a widget at random from the factory's widget inventory.
(a) What is the probability that the widget is defective?
(b) If the widget is defective, what is the chance that it came from Machine A?
Answer: a) 4+3+2=9%
b) P(A&B)= 9% * (1000/2000)
Not so quickly here.
I make a table
number built = 1000
% bad = .04
number bad = 40
number built = 600
number bad = 18
number built = 400
number bad = 8
Total built = 1000+600+400 = 2000
total bad = 40+18+8 = 66
66 out of 2000 are bad = .033 = 3.3 %
40 out of 66 bad are from A 40/66 = .606 = 60.6 %
I showed how to do #6 below scroll down
7. Suppose that there are nine adjacent parking spaces in one row of a parking lot. Nine cars are to be parked by an attendant. Three are SUVís, three are compacts, and three are expensive sports cars. Assuming that the attendant parks the cars randomly, find the chance that the three expensive sports cars are parked next to one another.
how many ways can the 3 blings be parked next to each other?
The first could be in space 1
or space 2
or space 3
------- continuing to
or space 7
but not 8 or 9
so 7 starting parking places.
now the three blings can be in various orders in those spaces
How many ways to order three cars in three spaces?
permutations of 3 vehicles taken one at a time
= 3!/2! = 6
so 7 * 6 = 42 ways
how many ways to park 9 cars any order?
9! I think
so I think the answer is
Good grief, I think I might take this course some time :)
thank you damon
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