What is the probability

  • Thread starter bhoover05
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Hey guys. I am having some issues understanding how to calculate these. If anyone could help, it would be greatly appreciated.

1. My friends came down to visit and wanted to go to a football game. Tickets were sold out, but I planned on asking my student's in lab throughout the day to see if they had an extra ticket I could buy. I had 3 friends coming down, plus I needed one myself. Thus I needed 4 tickets. If I had 125 students to ask, and each one had a probability of a 5% chance of having an extra ticket, what is the probability that my friends' and I will get to go to the game?

2. There are 5 red chips and 3 blue chips in a bowl. The red chips are numbered 1,2,3,4,5 and the blue chips are numbered 1,2,3. If two chips are drawn at random and without replacement, what is the probability that these chips have either the same number or the same color?

3. Machines I,II, and III are all producing springs of the same length. machines I,II,III produce 1%,4%, and 2% defective springs. Of the total production of springs in the factory, Machine I produces 30%, Machine II produces 25%, and Machine III produces 45%. If one spring is selected at random from the total springs produced in a given day, determine the probability that its defective.

4. a) Both of us roll a dice. If I have a higher number than you, I win. If not, you win. What are the chances that I win? b) What if, on a tie, we roll again. Now what are my chances of winning?

5. Game costs 5$ to play. If you decide to play, you will pull one card randomly from a deck of cards. If it is an Ace, you win. Otherwise you lose your $5. How much money would you have to win (for pulling an Ace) to make the game fair? How much should the prize money be to make this game worthwhile (to play a large number of times)?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi bhoover05! Welcome to PF! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :wink:
 
  • #3
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I am really awful at statistics, so I literally have to talk myself thru every step, but that will help everyone identify just what I am doing wrong!

1. So to find the probability that four will go to the game:
1-P(of not getting 4 tickets)
~keeping in mind the 5% chance of an extra ticket from each of the 125 students
so for 4:
1-P(3)-P(2)-P(1)=
(0.05)^3 x (0.95)^122
Here the 0.05 is from the 5% chance of an extra ticket for 3 tickets, and the 122 is from the 122 students which did not have a ticket. . .
Now I believe I continue this as:
(0.05)^2 x (0.95)^123
(0.05)^1 x (0.95)^124
But I am not quite sure. Do I obtain all of these values then plug them back into 1-P(3)-P(2)-P(1)?

Thank you oodles! I am going to rework thru the others to provide you with more mistakes to read through.
 
  • #4
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2. Both of us have a dice. (1/6 probability of rolling any number on the die)
Probability of pulling a red = 5/8
Probability of pulling blue= 3/8
Probability of having the same number OR same color?
Same number = (since 1-3 are the only similar numbers) 3/8
Same color= pulling a red= 5/8 or pulling blue= 3/8
I feel this should be more complicated that that.
 
  • #5
tiny-tim
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1. My friends came down to visit and wanted to go to a football game. Tickets were sold out, but I planned on asking my student's in lab throughout the day to see if they had an extra ticket I could buy. I had 3 friends coming down, plus I needed one myself. Thus I needed 4 tickets. If I had 125 students to ask, and each one had a probability of a 5% chance of having an extra ticket, what is the probability that my friends' and I will get to go to the game?

1. So to find the probability that four will go to the game:
1-P(of not getting 4 tickets)
~keeping in mind the 5% chance of an extra ticket from each of the 125 students
so for 4:
1-P(3)-P(2)-P(1)=
(0.05)^3 x (0.95)^122
Here the 0.05 is from the 5% chance of an extra ticket for 3 tickets, and the 122 is from the 122 students which did not have a ticket. . .
Now I believe I continue this as:
(0.05)^2 x (0.95)^123
(0.05)^1 x (0.95)^124
But I am not quite sure. Do I obtain all of these values then plug them back into 1-P(3)-P(2)-P(1)? .

Hi bhoover05! :smile:

You have half the general principle right …

P = 1-P(3)-P(2)-P(1)-P(0)

(I'll assume you correctly left out P(0) because it's neglligbly small! :wink:)

and eg p3(1-p)122 is the part of the correct formula for P(3) …

but that's only the probablity of Tom Dick and Sally (your favourite triple of students) of having an extra ticket …

you need to multiply by the number of "triples" of students, which is … ? :smile:
 
  • #6
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The number of triples? 125 students? Hm. . . would that be 125/3=41.7?
 
  • #7
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3. Total amount of defective springs = 1+4+2= 7%
If there is 7 percent of the total productive of springs that are defective, shouldn't the chances that a randomly defective spring also be 7%?

4. Dice= numbers 1-6.
Here is where i get confused. You could roll a two which could be higher if I roll a 1. Seems like there could be all sorts of combination's. Which is where the statics would come in.
a) So if 6 beats 1-5, 5 beats 1-4, 4 beats 1-3, 3 beats 1-2, and 2 beats 1. . .
1-P(5)-P(4)-P(3)-P(2)-P(1)
Would I have to use a random number table to get the probabilities of rolling each of those?
b) Roll again on a tie. Wouldn't the same probabilities for winning be the same as in part a) here since there is a new start or new roll?
 
  • #8
tiny-tim
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revise combinatorials

The number of triples? 125 students? Hm. . . would that be 125/3=41.7?

Nooo … it's the "combinatorial" 125C3

you really need to revise combinatorials before you go any further with these questions :frown:

(and i'm going out for the evening :smile:)
 
  • #9
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ooooh! thank you for your help!
 
  • #10
CRGreathouse
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2. Both of us have a dice. (1/6 probability of rolling any number on the die)
Probability of pulling a red = 5/8
Probability of pulling blue= 3/8
Probability of having the same number OR same color?
Same number = (since 1-3 are the only similar numbers) 3/8
Same color= pulling a red= 5/8 or pulling blue= 3/8
I feel this should be more complicated that that.

I think you're confusing problems 2 and 4. In this problem you're pulling two chips out of a bag, not rolling two dice. First, figure out the number of ways to pull:
1. Both red.
2. First red, then blue.
3. First blue, then red.
4. Both blue.

Remember, there are 8 chips in the bag for the first draw, but only 7 for the second draw.

#1 and #4 are 'automatic successes'. For #2 and #3, you have one of each color. Given that, what are the chances that both have the same number?
 
  • #11
CRGreathouse
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3. Total amount of defective springs = 1+4+2= 7%
If there is 7 percent of the total productive of springs that are defective, shouldn't the chances that a randomly defective spring also be 7%?

But the total isn't 7%. Consider this scenario:
Machine 1 produces 2% defective springs and machine 2 produces 50% defective springs. Machine 1 makes 100% of the springs in the factory; machine 2 produces 0%.

The percentage of defective springs (in that example) is 2%, not 52%.
 
  • #12
CRGreathouse
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4. Dice= numbers 1-6.
Here is where i get confused. You could roll a two which could be higher if I roll a 1. Seems like there could be all sorts of combination's. Which is where the statics would come in.
a) So if 6 beats 1-5, 5 beats 1-4, 4 beats 1-3, 3 beats 1-2, and 2 beats 1. . .

Let's make this easier. If your opponent rolls a 6, you have 0 ways to win. If your opponent rolls a 5, you have 1 ways to win. Add up all of these ways to win, then divide by the total number of outcomes (6 * 6 = 36).

b) Roll again on a tie. Wouldn't the same probabilities for winning be the same as in part a) here since there is a new start or new roll?

No. In (a), you lose on a tie; in (b), you reroll until the result isn't a tie.
 
  • #13
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I do believe that I have an idea as to how to do number 5.

E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited?
 
  • #14
tiny-tim
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#5

I do believe that I have an idea as to how to do number 5.

E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited?

Hi bhoover05! :smile:

Yes, that's right …

and for this problem, to "make the game fair", you need E(x) = 0 …

so amount gained = … ? :wink:
 
  • #15
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E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited

Alright. . .
~amount lost= $5
~prob. of loss= (number of cards that are not aces/full deck)= 48/52
~amount gained is what we are trying to figure out so we can determine the amount of prize money
~Probability of a win (number of aces/52 deck of cards)=(4/52)

I am having problems filling in what amount gained should be. . . I am trying to hard but none of my notes are helping.
 
  • #16
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bhoover05,

For #5 I believe the part that you're missing is that for the game to be fair, you want E(X)=0.

Edit: oops, didn't see that Tiny Tim had beaten me to the punch! Apologies.
 
Last edited:
  • #17
tiny-tim
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E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited

Alright. . .
~amount lost= $5
~prob. of loss= (number of cards that are not aces/full deck)= 48/52
~amount gained is what we are trying to figure out so we can determine the amount of prize money
~Probability of a win (number of aces/52 deck of cards)=(4/52)

I am having problems filling in what amount gained should be. . . I am trying to hard but none of my notes are helping.

Hi bhoover05! :smile:

Yes … chance of win = 1/13, chance of loss = 12/13

So be systematic: if X(lose) = -5 and X(win) = W,

what is E(X)? :smile:
Edit: oops, didn't see that Tiny Tim had beaten me to the punch! Apologies.

confirmation is always useful :biggrin:

i think sometimes members aren't convinced that the first answer is right! :wink:
 
  • #18
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So for a fair game, E(x)=0

E(x)= (amount lossed)(prob. of loss)+(amount gained)(prob. of a win)
where x= expected value of the profited

Alright. . .
~amount lost= $5, SO -5
~prob. of loss= (number of cards that are not aces/full deck)= (48/52) OR (12/13)

~amount gained (Win)= x(win) or W
~Probability of a win (number of aces/52 deck of cards)=(4/52) OR (1/13)

SO (drum roll)
E(x)= (-5)(12/13)+(W)(1/13)
= -4.6153 +0.0769W
Would this then be a divide -4.6153 by 0.0769W to get 60.49?!
 
  • #19
tiny-tim
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SO (drum roll)

drum-roll-on-the-floor!! :rofl: :rofl: :rofl: :rofl:
E(x)= (-5)(12/13)+(W)(1/13)
= -4.6153 +0.0769W
Would this then be a divide -4.6153 by 0.0769W to get 60.49?!

yes, but … d'oh!

are you deliberately making this complicated to impress the audience? :rolleyes:

0 = E(x)= (-5)(12/13)+(W)(1/13),

so (multiplying by 13), W = … ? :biggrin:
 
  • #20
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0 = E(x)= (-5)(12/13)+(W)(1/13),

so, W = 13W?
Meaning -4.6153 should be divided by 13W?

And no! I am really trying to learn this. . . I greatly appreciate your help! I am a psychology major with an emphasis with behavior analysis- I never work with stats on my day to day life! But I have been reading and trying to up my game!
 
  • #21
tiny-tim
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0 = E(x)= (-5)(12/13)+(W)(1/13),

so, W = 13W?

No …

0 = E(x)= (-5)(12/13)+(W)(1/13),

multiply by 13 …

0 = (-5)(12) + (W)(1) …

so … :smile:
 
  • #22
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AH! :-)!

(drum roll. . . )

0 = (-5)(12) + (W)(1) …
60= 1W
W= 60 EVEN :-) !

(right*crosses fingers*- third time is a charm???)
 
  • #23
tiny-tim
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taraaa …

AH! :-)!

(drum roll. . . )

0 = (-5)(12) + (W)(1) …
60= 1W
W= 60 EVEN :-) !

(right*crosses fingers*- third time is a charm???)

Yeeees!! :biggrin:

And now see why it works :wink:

with something easy like this, you should always be able to express it in English …

in this case, you're going to lose 12 times as often as you're going to win …

so you need 12 times the payout to break even! :smile:
 
  • #24
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For the question about the friends coming down to attend a football game, I do believe I have to use the equation. . .

(n/x)p^x(1-p)^n-x

Here, I think, you would use 125 for n (since there are 125 students asked), 4 for x (since there is a total of four tickets needed), and the p would be 0.05 (since there is a 5% chance of each one having an extra ticket)?
Or more simply:
n=125
x=4
p=0.05

(n/x)p^x(1-p)^n-x
SO:
(125/4)(0.05)^(4) (1-(0.05))^(125)-4= Probability that all four will get to go to the game

SO: Probability that all 4 can attend= 0.122?
 
  • #25
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Yes, I see how that works. I can NOT believe I was so easily confused by some basic algebra! AH! I feel quite ridiculous having stumbled through that the way I had. Thank you so much for your support!
 

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