What are the Probabilities in These Different Scenarios?

[bhoover05]

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)?

 

[tiny-tim]

Welcome to PF!

Hi bhoover05! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[bhoover05]

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.

 

[bhoover05]

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.

 

[tiny-tim]

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!

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! )

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 … ?

 

[bhoover05]

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

 

[bhoover05]

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?

 

[tiny-tim]

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

(and I'm going out for the evening )

 

[bhoover05]

ooooh! thank you for your help!

 

[CRGreathouse]

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?

 

[CRGreathouse]

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%.

 

[CRGreathouse]

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.

 

[bhoover05]

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?

 

[tiny-tim]

#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!

Yes, that's right …

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

so amount gained = … ?

 

[bhoover05]

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.

 

[rochfor1]

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.

 

[tiny-tim]

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!

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)?

Edit: oops, didn't see that Tiny Tim had beaten me to the punch! Apologies.

confirmation is always useful …

i think sometimes members aren't convinced that the first answer is right! ​

 

[bhoover05]

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?!

 

[tiny-tim]

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?

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

so (multiplying by 13), W = … ?

 

[bhoover05]

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!

 

[tiny-tim]

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 …

 

[bhoover05]

AH! :-)!

(drum roll. . . )

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

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

 

[tiny-tim]

taraaa …

AH! :-)!

(drum roll. . . )

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

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

Yeeees!

And now see why it works …

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! ​

 

[bhoover05]

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?

 

[bhoover05]

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!

 

[tiny-tim]

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!

hee hee!

For the question about the friends coming down to attend a football game, I do believe I have to use the equation. . .

There's no simple way of doing this …

the number of spare tickets wil be between 0 and 125 …

you want the probability that it's 4 or 5 or … or 125.

So it's best to calculate the probability that it's 0 or 1 or 2 or 3, and then subtract that from 1.

So, to start you off … what's the probability that it's 0, or 1?

 

[bhoover05]

Here I was, thinking I was on a roll. . . Ugh.

Well, would this involve using the same formula? The:

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

the p would still equal 0.05 since its 5%?

 

[bhoover05]

^ and the n would still be 125 since there are 125 students. . .

 

[bhoover05]

I have ran thru a couple of other problems that are a lot easier to see if I have improved at all. . . Thought maybe you would enjoy checking over them :-)?

1. A game of Yahtzee, you get 3 rolls of the dice per turn, and you get to choose which ones you re-roll each time. My brother calls me up and asks me the probability of him winning the game. If he gets a large straight, he wins, otherwise my Dad wins. With his first roll, he got a 2,3,4,6,6. He is going to re-roll the second 6 that he got, to try and get a 5. He has two rolls to get a 5. Whats the probability that he gets that 5 and beats my dad?
~~1st roll: fail to get a 5 w/ probability of (5/6)
probability to fail 2x's=(5/6)^2=0.694
Probability = 1-0.694= 0.3055556
Probability= 0.3056

2. You and I are playing a game. I'm going to be flipping a coin, and you will be rolling a die. I get to go first. If I get heads, I win the game, but if you roll the die and get the number 1,2,3,4, you win. It goes back and forth until someone wins. Who has the better chance of winning and why?
~~You automatically have a 0.5, or 50% chance of winning your coin toss on heads
My roll only takes place if you roll a tail
Your second flip only takes place if I lose my roll ( on a 5 or 6, thus a 1/3 probability)
Probability= E (1/2)^i (1/3)^i-1
(1/2)=coin
(1/3)=die
Probability for coin tossing player (you)= 0.6
Me= (rolling player) = 1-p= (1-0.6)= 0.4
My probability= 0.4 vs you with the 0.6 probability

3. (same set up as question above) BUT, I go first. If I get heads, I win. If not, you roll the die, and if you get the numbers 1,2,3,4,5 you win. If not, back to me, and if I get heads, I win. If not, you win by default. Who has a better chance of winning, and why?
~~P=E (1/2)^i (5/6)^i-1
(1/2)=coin
(5/6)=die
Assuming there is a 50% chance of winning going first, you still have a better chance of winning the game overall.

4. After a game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt a arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.
~~For this one I drew a diagram and tried to assign the injuries to the players. When putting the knee injuries on the diagram, they would have to be overlaped to make the data work. This means that there would be injuries than listed, thus meaning the data is impossible.

5. Player A and B play a sequence of independent games. Player A throws a die first and wins on a 6. If he fails, player B throws and wins on a five or six. If he fails, A throws and wins on a four, five, or six. ( this continues in same pattern) Find the probability of each player winning.
~~A starts: win= (1/6)
lose=(5/6)
(1/6)+((5/6)*(4/6)*(3/6))+((5/6)*(4/6)*(3/6)*(2/6)*(1/6))
- Player A has a 52% probability
~~B goes second: win=(2/6)
lose=(4/6)
- Player B has a 48% chance

6. Five people are about to play Yahtzee. Whats the probability that I win the game?
~~ Since there are 5 people playing, there is a 1/5 chance of winning the game.