Saturday

January 21, 2017
Posted by **Amelie** on Sunday, February 17, 2008 at 3:54pm.

(a) no note can be used more than once?

(b) any note can be used as often as you please?

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Ans for a) a permutation

12P8 =19,958,400

b) a combination

12C8= 495

Is this correct?

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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?

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Answer: P(A&B)=(.45)(.50)= 22.5%

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4. In a certain election, the incumbent Republican will run against the Democratic nominee. There are three Democratic candidates, D1, D2 and D3, whose chances of gaining the Democratic nomination are .50, .35 and .15, respectively. Here are the chances that the Republican will win against each of these possible Democratic nominees:

vs. D1: 0.60 vs. D2: 0.50 vs. D3: 0.40

(a) Name (but do not give) the probability formula that is needed to find the chance that the Republican will win the election.

(b) Find the probability that the Republican will win the election.

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not sure how to do this one.

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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?

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Answer: a) 4+3+2=9%

b) P(A&B)= 9% * (1000/2000)

= 4.5%

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6. A fair coin is flipped three times. You win $5 every time the outcome is heads. Let the random variable X represent the total number of dollars you win.

(a) List the sample space.

(b) Determine the probability function of X.

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Answer: a) 2^3= 8 possibilities

b) (.5+.5+.5)/15= 1/10

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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 SUV’s, 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.

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Answer: not sure how to do this one.

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- Probability -
**Amelie**, Sunday, February 17, 2008 at 3:55pmposted twice, sorry

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**Damon**, Sunday, February 17, 2008 at 4:42pmI di #4 already, scroll down

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**Damon**, Sunday, February 17, 2008 at 5:02pm1.) 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?

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Ans for a) a permutation

12P8 =19,958,400

b) a combination

12C8= 495

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I believe that your answer for b

"a combination

12C8= 495"

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

12^8

which is very big indeed ! - Probability -
**Damon**, Sunday, February 17, 2008 at 5:14pm3. 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?

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Answer: P(A&B)=(.45)(.50)= 22.5%

+++++++++++++++++++++++++++++++++++

hmm

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 - Probability -
**Damon**, Sunday, February 17, 2008 at 5:28pm5. 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?

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Answer: a) 4+3+2=9%

b) P(A&B)= 9% * (1000/2000)

= 4.5%

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Not so quickly here.

I make a table

Machine A:

number built = 1000

% bad = .04

number bad = 40

Machine B:

number built = 600

number bad = 18

Machine C:

number built = 400

number bad = 8

Total built = 1000+600+400 = 2000

total bad = 40+18+8 = 66

SO:

66 out of 2000 are bad = .033 = 3.3 %

40 out of 66 bad are from A 40/66 = .606 = 60.6 % - Probability -
**Damon**, Sunday, February 17, 2008 at 5:29pmI showed how to do #6 below scroll down

- Probability -
**Damon**, Sunday, February 17, 2008 at 5:48pm7. 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 SUV’s, 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.

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++++++++++=++++++++++++++++++++++++++++

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

42/9! - Probability -
**Damon**, Sunday, February 17, 2008 at 5:50pmGood grief, I think I might take this course some time :)

- Probability -
**Amelie**, Sunday, February 17, 2008 at 10:34pmthank you damon