All Answers

+2354

Gtorres:
What is 5+4y< 5 ?

CodeCogsEqn (1).gif

if you have trouble solving inequalities look at this page

http://www.mathsisfun.com/algebra/inequality-solving.html

reinout-gMar 5, 2014

Alan:
PabloDons:
Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

Your question has more than one interpretation!

If you mean: "What is the probability of getting at least one 6 in six throws of the dice" then this is just 1 - the probability of getting no sixes.
The probability of getting no six for the first die is 5/6.
The probability of getting no six for either the first or second die is (5/6)*(5/6) or 5 ²/6 ²
The probability of getting no six for any of the first three dice is (5/6)*(5/6)*(5/6) or 5 ³/6 ³
Continuing this way, the probability of getting no sixes for any of the six dice is 5 ⁶/6 ⁶
Hence, the probability of getting at least one six is 1- 5 ⁶/6 ⁶

1-5^6/6^6

If you mean "What is the probability of getting exactly one 6 in six throws of the dice" then
The probability of one particular die getting a 6 and all the others getting no sixes is (1/6)*(5 ⁵/6 ⁵)
But there are 6 dice each of which could be the one that falls as a 6, so the probability is 6*(1/6)*(5 ⁵/6 ⁵) or 5 ⁵/6 ⁵

5^5/6^5

If you mean something else then you'll need to specify exactly what!

reinout-g:

PabloDons:
Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

There are two ways in which I can interpret your question, You're either asking what the odds are of throwing one six in six throws, or you're asking the odds of throwing at least one six in six throws.

The odds of throwing one six in six throws can be calculated in the following way;

We know that there is a (1/6)th chance of throwing a six with one dice and a chance of (5/6) of not throwing a six with one dice.

So the odds of throwing for example [6] [no 6] [no 6] [no 6] [no 6] [no 6] = (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)

However this is only the chance of throwing a six in the first throw and no six in the next 5 throws.

Another possibility would be throwing [no 6] [6] [no 6] [no 6] [no 6] [no 6] or [no 6] [no 6] [no 6] [no 6] [no 6] [6]

All these possibilities have the same probability as the one I just calculated ( (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) )

There are 6 possibilities for which in six throws you throw 1 six (in the first, second, third, fourth, fifth, or sixth throw)

So the odds of throwing 1 six in six throws is 6 * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 0.4019 ( = 40.19%)

These odds take into account the cases where there is only one six, and it leaves out the cases where you throw six twice, three times, or even more times

To calculate the odds where you want at least one six you need to make use of the following;

The chance of something happening = 1 - the chance of that thing NOT happening

So for example if I have 0.30 (=30%) chance of winning $1000, I have 0.70 (=70%) chance of NOT winning $1000.

Similarly, to calculate the probability of throwing at least 1 six, It is more convenient to calculate the probability of throwing less than 1 six.
Or more clearly, the probability of throwing no sixes at all.

The probability of throwing no sixes means that each dice hits the (5/6)th chance of NOT throwing a six.
Consequently the probability of NOT throwing a six = (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) = 0.3349 (or 33.49%)

Now to find the odds of throwing at least one six, we simply need to calculate 1-0.3349 = 0.6651

So there is a 66.51% chance of throwing at least one six.

As you can see, the formulation of the question is very important in calculating the right probability

Thank you very much, I was thinking of the odds to get at least 1 six out of 6 throws. Thank you rainout for the brilliant example of a the chance for something to happen = 1 - the chance of something not to happen. I didn't think of that at all.

GuestMar 5, 2014

+2354

PabloDons:
Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

There are two ways in which I can interpret your question, You're either asking what the odds are of throwing one six in six throws, or you're asking the odds of throwing at least one six in six throws.

The odds of throwing one six in six throws can be calculated in the following way;

We know that there is a (1/6)th chance of throwing a six with one dice and a chance of (5/6) of not throwing a six with one dice.

So the odds of throwing for example [6] [no 6] [no 6] [no 6] [no 6] [no 6] = (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)

However this is only the chance of throwing a six in the first throw and no six in the next 5 throws.

Another possibility would be throwing [no 6] [6] [no 6] [no 6] [no 6] [no 6] or [no 6] [no 6] [no 6] [no 6] [no 6] [6]

All these possibilities have the same probability as the one I just calculated ( (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) )

There are 6 possibilities for which in six throws you throw 1 six (in the first, second, third, fourth, fifth, or sixth throw)

So the odds of throwing 1 six in six throws is 6 * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 0.4019 ( = 40.19%)

These odds take into account the cases where there is only one six, and it leaves out the cases where you throw six twice, three times, or even more times

To calculate the odds where you want at least one six you need to make use of the following;

The chance of something happening = 1 - the chance of that thing NOT happening

So for example if I have 0.30 (=30%) chance of winning $1000, I have 0.70 (=70%) chance of NOT winning $1000.

Similarly, to calculate the probability of throwing at least 1 six, It is more convenient to calculate the probability of throwing less than 1 six.
Or more clearly, the probability of throwing no sixes at all.

The probability of throwing no sixes means that each dice hits the (5/6)th chance of NOT throwing a six.
Consequently the probability of NOT throwing a six = (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) = 0.3349 (or 33.49%)

Now to find the odds of throwing at least one six, we simply need to calculate 1-0.3349 = 0.6651

So there is a 66.51% chance of throwing at least one six.

As you can see, the formulation of the question is very important in calculating the right probability

reinout-gMar 5, 2014

+33666

PabloDons:
Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

Your question has more than one interpretation!

If you mean: "What is the probability of getting at least one 6 in six throws of the dice" then this is just 1 - the probability of getting no sixes.
The probability of getting no six for the first die is 5/6.
The probability of getting no six for either the first or second die is (5/6)*(5/6) or 5 ²/6 ²
The probability of getting no six for any of the first three dice is (5/6)*(5/6)*(5/6) or 5 ³/6 ³
Continuing this way, the probability of getting no sixes for any of the six dice is 5 ⁶/6 ⁶
Hence, the probability of getting at least one six is 1- 5 ⁶/6 ⁶

1-5^6/6^6

If you mean "What is the probability of getting exactly one 6 in six throws of the dice" then
The probability of one particular die getting a 6 and all the others getting no sixes is (1/6)*(5 ⁵/6 ⁵)
But there are 6 dice each of which could be the one that falls as a 6, so the probability is 6*(1/6)*(5 ⁵/6 ⁵) or 5 ⁵/6 ⁵

5^5/6^5

If you mean something else then you'll need to specify exactly what!

AlanMar 5, 2014

+2354

Melody:
*
Would like to comment on, or add to anything that I have said regarding communication with Andrew Massow.
If so, this would be an appropriate place to have your say.

Melody.

If mr. Massow would embed this in the forum we would all be able to use LateX with the forum...

http://www.codecogs.com/latex/quick-install.php

http://www.codecogs.com/pages/forums/pagegen.php?id=1682

reinout-gMar 5, 2014

#23

+2354

And here is the new puzzle!
If you have trouble solving it, feel free to give the easier one a try.

Easier one: (still to be solved)

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one :

Three kids named Brooks, Clint and Riley have a new game using water balloons.
They each take turns trying to throw one balloon at one of the other two. If they hit, the person is out, otherwise the next person may take his turn.

Brooks hits 1/3 of the time, Clint hits 2/3 of the time and Riley never misses.

Brooks is allowed to start, then Clint (given that he is still in the game), then Riley (given that he is still in the game), etc.

Where should brooks aim during his first turn?

reinout-gMar 5, 2014

+201

God morning. Thank you much for reply. I got it.

mathcalcMar 5, 2014

#22

+2354

+11

Melody:
reinout-g:
Since I do not have time for a new puzzle, I'll post something which people can comment on.

Steve flips a coin,
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail.

How much would you be willing to pay to play this game?

There is no talk of losing any money here. Does it cost something to play? If it doesn't cost anything and the coin is 'fair' (I can't think of the proper word) Then I would play until i got too too bored, or until I won enough money for that trip to The Netherlands that i have always desired. .
I didn't read it properly - what a shame.

If I played $1 each game, eventually I would have to win.
If I paid $2 per flip then I would break even if tails came up on the first toss and win otherwise. So long as i keep putting my money down.
If i paid $4 per flip then I would only lose money if tails came up on the 1st toss, I would break even if tails came second and after that I would win.
So, it doesn't matter how much I pay (assuming each toss costs the same). so long as I get the chance to play again after each win, I cannot lose (eventually)
NOW you can tell me that I misinterpreted the question again (I always have a problem following the dots) ?

I posted this one a little differently because it is more of a philosophical probability question than a riddle. It points out the flaw in expected value.

I like your idea of missing a number of times one can play the game, however people also buy lottery tickets which is pretty much the same for the less probable outcomes.

Actually, you could rephrase the question to a single lottery ticket where the odds of winning are equally distributed (so E(x) = (1/2) * $2 + (1/4) * $4 + (1/8) * $8.... => infinity)

I'll post a riddle (no paradox this time) somewhat later today...

reinout-gMar 5, 2014

#16

+2354

Melody:
reinout-g:
New puzzle!
Three rivalling coaches (Pierre, Pieter and Peter) and their three athletes (Édouard, Eduard and Edward) want to cross a river.
There is only one boat which can carry two persons.
The coaches do not want other coaches to be with their athlete without they themselves beeing present at any moment..
How can they cross the river?

Firstly, i can't handle all those Ps and Es
Mine are going to be Cs and As

We have A1, C1, A2, C2, A3, C3 to get across the river
A1 and C1 paddle across and A1 stays

We have C1, A2, C2, A3, C3 to get across the river
A2 and C2 paddle across and A1, A2 stay

We have C1, C2, A3, C3 to get across the river
A3 and C3 paddle across and A1,A2 and A3 stays

We have C1, C2, C3 to get across the river
C1 and C2 paddle across and A1, C1, A2, C2 stay A3 returns (he doesn't talk to anyone first - is that alright?)

C3 and A3 paddle across and then they are all there.

Is that okay or is there a better answer?

NO WE HAVE TO SCRAP THIS ONE IT DOESN'T FOLLOW THE RULES - SORRY.

You're on the right track,
but indeed, your answer is not correct.

try again

reinout-gMar 5, 2014

#58

+118728

+14

Hi all,
Another busy day today. Lots of fabulous answers from reinout-g, Alan, scrutinizer, Statman74, Dms, Alexandra and Rom. Thank you so much.
You know, when I wake in the morning and find all the night questions already answered it makes my day so much more pleasant. Thank you.
Monkey Spanker also answered one but an answer is no help if it doesn't come with some sort of an explanation. I'm sorry but it is the truth.
Both I and Superextremebro tried our hand at one of Reinout's puzzles. I know mine is wrong. I already had it up before I realized so I left it there.
Anony gave a nice thank you too. We always appreciate good manners.

Now, Dms dropped in and hugely impressed Reiner and me. He implanted LaTex into the forum and we want a lesson (or 2) on how it is done.
Please Dms, won't you come back and show us?
If you would like to take a look, this is the thread.
http://web2.0calc.com/questions/math_1546

I nearly forgot. I sent the email off to Andre Massow. I have already briefed you on its content.

That is it for my evening.
Enjoy the rest of your day.
Melody.

MelodyMar 5, 2014

+118728

bump

MelodyMar 5, 2014

+118728

Alan:
ABDULLAH:
If $10,000 is invested at an interest rate of 2% per year, compounded semiannually, find the value of the investment after the given number of year. 4 year

The formula for calculating compound interest is F = P(1 + r/n) ^nt where P is the amount invested ($10,000), r is the interest rate expressed as a fraction (0.02), n is the number of times compounded in a year (2), t is the number of years invested (4) and F is the future value (i.e. the amount the investment is worth after t years). Have a go at the calculation now you have this information.

Thanks Alan.

I use a different version of the same formula

S = P(1+i) ⁿ

Where p is the principal = $10000
i is the interest r**e per compounding period = 0.02/2 = 0.01
n is the number of compounding periods = 4*2=8

S = 10000(1.01) ⁸

10000*1.01^8=
$10828.57

MelodyMar 5, 2014

+33666

ABDULLAH:
If $10,000 is invested at an interest rate of 2% per year, compounded semiannually, find the value of the investment after the given number of year. 4 year

The formula for calculating compound interest is F = P(1 + r/n) ^nt where P is the amount invested ($10,000), r is the interest rate expressed as a fraction (0.02), n is the number of times compounded in a year (2), t is the number of years invested (4) and F is the future value (i.e. the amount the investment is worth after t years). Have a go at the calculation now you have this information.

AlanMar 5, 2014

#21

+118728

+11

superextremebro:
just switch to the other box so you get a 66 percent chance instead of a 33 percent chance.

What other box. What question are you answering?

When a thread is long, it is always a good idea to include the question with the answer.

reinout-g will probably know what you are talking about. This thread is his baby ( I hope that sounds positive ).

MelodyMar 5, 2014

+118728

jilliennek:
solve equation for x :
2x^2 + 9x = -5

We have all been pretty wowed by Dms's solution.
We have never seen LaTex in here like this before and we would love to know how it is done.
Please show us Dms.

However, I think Dms may have made the problem more difficult than it really is ('a' does not need to be 1)
quadratic formula.JPG

2x ² + 9x + 5 = 0

a=2, b=9 c=5

x=[ -9 ± sqrt( 81-40) ] / 4

x=[ -9 ± sqrt( 41) ] / 4

MelodyMar 5, 2014

#15

+118728

reinout-g:
New puzzle!
Three rivalling coaches (Pierre, Pieter and Peter) and their three athletes (Édouard, Eduard and Edward) want to cross a river.
There is only one boat which can carry two persons.
The coaches do not want other coaches to be with their athlete without they themselves beeing present at any moment..
How can they cross the river?

Firstly, i can't handle all those Ps and Es
Mine are going to be Cs and As

We have A1, C1, A2, C2, A3, C3 to get across the river
A1 and C1 paddle across and A1 stays

We have C1, A2, C2, A3, C3 to get across the river
A2 and C2 paddle across and A1, A2 stay

We have C1, C2, A3, C3 to get across the river
A3 and C3 paddle across and A1,A2 and A3 stays

We have C1, C2, C3 to get across the river
C1 and C2 paddle across and A1, C1, A2, C2 stay A3 returns (he doesn't talk to anyone first - is that alright?)

C3 and A3 paddle across and then they are all there.

Is that okay or is there a better answer?

NO WE HAVE TO SCRAP THIS ONE IT DOESN'T FOLLOW THE RULES - SORRY.

MelodyMar 5, 2014

#20

just switch to the other box so you get a 66 percent chance instead of a 33 percent chance.

GuestMar 5, 2014

+118728

briano15:
:? 6:15.17-6:10.76

6 : 15 . 17-
6 : 10 . 76
--------------
They both have 6 minutes so the question is really just

15.17 - 10.76 seconds = 4.41 seconds

15.17-10.76=

MelodyMar 5, 2014

+33666

bookstar:
ln(e^-9x)

ln(e ^-9x) is just -9x. This is because ln is the natural or Napierian logarithm - that is, the logarithm to the base e. In general log _b(b ^a) = a (where log _b is logarithm to the base b). In your example, b is e and a is -9x.

AlanMar 5, 2014

+118728

mathcalc:
Kenneth has $5. He spends g cents everyday. How much money does he have left after one week?

a. Express your answer in cents.
b. Express your answer in dollars.