little help with hw

[fonz87]

little stuck on these problems, any help would be awesome.

Two cards are dealt off the top of a well shuffled deck. You have a choice:

(a) To win $1 if the first is a king.

(b) To win $1 if the first is a king and the second a queen.



Which option is better? Or are they equivalent?

For credit, justify by showing what is the probability in each case.



problem 2

Four cards are dealt off the top of a well shuffled deck. You have a choice:

(a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.

(b) To win $1 if the four cards are of four different suits, in any order.



Which option is better? Or are they equivalent?

For credit, justify by showing what is the probability in each case.




problem 3

A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need

(a) no repairs at all?

(b) No more than one rpair? (hint: this is the same as: probability of one repair or no repair).

(c) Some repairs? (hint: this is the same as: probability of once, or twice, or three & more)



thanks

 

[flight doc89]

little stuck on these problems, any help would be awesome.

Two cards are dealt off the top of a well shuffled deck. You have a choice:

(a) To win $1 if the first is a king.

(b) To win $1 if the first is a king and the second a queen.



Which option is better? Or are they equivalent?

For credit, justify by showing what is the probability in each case.

Answer is A: Probability = (4/52)
4 kings, 52 cards

reason why A: for (b), probability is (4/52)*(4/51)
4 kings, 52 cards, after that first card is drawn, still 4 queens, but only 51 cards


problem 2

Four cards are dealt off the top of a well shuffled deck. You have a choice:

(a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.

(b) To win $1 if the four cards are of four different suits, in any order.


Which option is better? Or are they equivalent?

For credit, justify by showing what is the probability in each case.

Probability of (a) and (b) = (13/52)*(13/51)*(13/50)*(13/49)




problem 3

A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. What is the probability that a car chosen at random will need

(a) no repairs at all?

(b) No more than one repair? (hint: this is the same as: probability of one repair or no repair).

(c) Some repairs? (hint: this is the same as: probability of once, or twice, or three & more)

Is that the exact wording? Does your teacher like trick questions? I ask because none of the questions have a time-frame.

Assuming it's not a trick question:

(a) = (72/100)

(b) = (89/100) or (72/100)+(17/100)

(c) = (28/100) or (17/100)+(7/100)+(4/100)

thanks

reply in bold. if i made a mistake somewhere, someone correct me.

 

[Quin]

1- A, 1/13 vs (1/13)*(4/51)
2- B, don't feel like figuring out probabilities here.
3- A) 18/25
B) 89/100
C) 7/25

I say B on 2 because in B the first pull is a guarantee, in A it must be a club and therefore is a 1/4 instead of a 1.

EDIT: I'm a dumb ass there are only 13 cards in a suit lol

 

[fonz87]

Canada has two official languages, English and French. Choose a Canadian at random and ask, “What is your mother tongue?” Here is the distribution of responses, combining mny separate languages from the broad Asian/Pacific region:

Language

Probability

English
0.59

French

0.23

Asian/Pacific
0.07

Other
?


(a) What probability should replace “?” in the distribution?

(b) What is the probability that a Canadian’s mother tongue is not English?

(c) What is the probability that a Canadian’s mother tongue could be either English or French?


just wondering if i did this right

A - .59+.23+.07 = .89 .89/3 = .296....7

B - .59+.23+.07+.29 = 1.18 ...... so .59/1.18 = .50

C - .59+.23 = .82 .82/1.18 = .69


Did i do this problem right ??

 

[Quin]

No. I don't understand why you did any of that, actually.
A) .11
B) .41
C) .82

 

[fonz87]

how did you get the .41 ?? and the .82 ?

thanks for the help guys!!

 

[Quin]

1-.59, .59+.23. You're over thinking it. Np

 

[Sil]

Aren't these binomial and geometric probabilities? I should know as I just took the exam for it last week, but I'm too damn lazy to go look in my notes for the plug and play formulas.

 

[flight doc89]

I say B on 2 because in B the first pull is a guarantee, in A it must be a club and therefore is a 1/4 instead of a 1.

You are right, i figured it wrong.

(a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.
Probability of (a) = (13/52)*(13/51)*(13/50)*(13/49)

(b) To win $1 if the four cards are of four different suits, in any order.

Since it doesn't matter what the first card is, it becomes (52/52)*(13/51)*(13/50)*(13/49)

answer is B

 

[fonz87]

anyone know what this is asking for
im so lost on this question


“You want to determine the best color for attracting cereal leaf beetles to boards on which they will be trapped. You will compare four colors: blue, green, white, and yellow. The response variable is the count of beetles trapped. You will mount one board on each of 16 poles evenly spaced in a square field, with four poles in each of four rows. Sketch the field with the locations of the 16 poles. Outline the design of a completely randomized experiment to compare the colors. Randomly assign colors to the poles, and mark on your sketch the color assigned to each pole. Use table B and start at line 115

 

[Quin]

Paypal me cash and I'll just do it all for you

lol

 

[fonz87]

paypal ?? and how much ?? 5 ok ?

 

[Quin]

L-O-L

I just added this bit so I could keep my lawlin' capitalized!

 

[fonz87]

so how is that problem done ?? its confusing me a little ??
any help would be great.

Thanks for all the helps guys... it really helped me

 

[dugums]

You are right, i figured it wrong.

(a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order.
Probability of (a) = (13/52)*(13/51)*(13/50)*(13/49)

(b) To win $1 if the four cards are of four different suits, in any order.

Since it doesn't matter what the first card is, it becomes (52/52)*(13/51)*(13/50)*(13/49)

answer is B

Answer is B, but your probabilities are still wrong.

First pull: 52/52
Second pull: 39/51
Third pull: 26/50
fourth pull: 13/49

 

[fonz87]

so anyone know how to do that last problem with the colors.. no idea whats it is asking ?

 

[Facime]

do your own homework!


...and we wonder what is wrong with the educational system?

 

[fonz87]

damn i asked for a little help, i guess it's too much to ask now.. huh

hahah

 

[Sil]

so anyone know how to do that last problem with the colors.. no idea whats it is asking ?

You're basically designing a completely randomized experiment using whatever random digit table you were provided, number the 16 poles 0-15, or 1-16 or realistically anything you want, then assign a set of digits to each color, go through your table and assign each pole a color. Then put it in a sketch.

 

[dugums]

so anyone know how to do that last problem with the colors.. no idea whats it is asking ?

The problem is actually giving you step-by-step instructions of what to do. Just follow them. If you don't get it, read it again. Don't over-think these things. If you don't know what it's asking, just start working on it.

Post your work and we'll correct you.

You aren't going to learn anything by us posting the answers.