Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

hi Hannibal lecter

A university committee should have 5 members. There are 5 candidates from the computer department, and 10 candidates from the math dept.

a) Find the probability of event that 2 members would be from computer and 3 from math.

b) Is there a possibility that all the members are from math

c) Find the possibility that there is at least 1 person from computer

For (a) you need to calculate the probability of 2 from 5 (computer) and the probability of 3 from 10 (mathematics) and then put these together into a single probability.

I'll start you off with this:

P(2 from computers) = 5 / 15 x 4 / 14

But the same result can occur in 2 ways: eg. C1 followed by C2 or C2 followed by C1. So you're in danger of counting the same result twice. We avoid this by dividing by 2

P(2 from computers) = 5 / 15 x 4 / 14 ÷ 2

Then for the maths members you start with 10 / 13 and then continue for two more members. This time to avoid repetitions you need to divide by 6 as

M1M2M3, M1M3M2, M2M1M3, M2M3M1, M3M1M2, M3M2M1 all give the same three maths members.

Please have a go at finishing this off and post your working. I'll check it for you.

Bob

hi Hannibal lecter

"3 in a row" means you throw a coin once and it's a head, then throw again and it's a head and then again it's a head for a third time.

In a probability question where there is only one event the formula is

probability = (number of ways you get the result you're interested in) / ( total number of things that could happen)

In the case of throwing a coin and getting a head, there are two possibilities (H or T) and one of these gives what you want, so P = 1/2 = 0.5

In situations where more than one event occurs the formula for calculating all the things happening is

probability = (probability of event one) x (probability of event two) x (probability of event three) x ..........

So for three heads that is 0.5 x 0.5 x 0.5

Each coin cannot 'know' what happened with the other coins, so for each the probability is 0.5.

Here's another example:

I pick three cards from a usual pack of 52. What is the probability of getting three aces ?

prob(first ace) = 4/52

But now that ace has gone so

prob(second ace) = 3/51

and now there are only 2 aces left so

prob(third ace) = 2/50

So Prob(three aces) = 4/52 x 3/51 x 2/50 = 0.000181

Bob

hi Hannibal lecter

what is the probability of finding 2 people have the same clothes out of 50 poeple???

With so little information this is not a question that a mathematician can answer.

Suppose the 50 are soldiers on parade. They should all have the same clothes so the probability in this case is very nearly 1 (we have to allow the small possibility that a soldier may not have the correct uniform)

If the 50 are people who hate being seen wearing the same as any of the others, then the probability will be close to zero.

To make a sensible maths question more information is required.

Bob

hi Hannibal lecter

Hi, this is a part of population grew

2003 12.853

2004 13.290Difference is 1.34

not 0.34

To work out the percentage growth you need this formula:

% increase = (new value - old value)/old value x 100

new value - old value = 13.29 - 12.853 = 0.437

Please check your working here as you have an incorrect value.

so % increase = 0.437/12.853 x 100 = 3.3999984 which is about 3.4%

Bob

Try drawing a diagram. I think this will show you what is happening.

Bob

Hi Hannibal lecter,

As you know two points you can get the equation of the straight line in the form y = mx + d. Substitute that value of y into y = x squared to get a quadratic in x. It should be easy to factorise as you know (x-1) is one factor. The other will give the x coordinate of C.

Bob

Hi Goutam,

Welcome to the forum.

The wage bill will go down by a multiplier of times 7/11 and up by times 13/10

Bob

Hi

Do you have formulas for sinh(x + y) and cosh(x+y) ?

If yes, use tanh = sinh / cosh

Or go back to definitions in terms of "e"

Bob

hi Monox D. I-Fly

I calculated a = -9 as well so I think that part is correct.

It'll be difficult to show a division on the forum so I'll try an alternative approach. It amounts to the same thing but will allow me to proceed one step at a time.

2x^3-5x^2 -9x+18 is to be divided by (x-1) so the result will be a quadratic. By inspection the x^2 term must be 2x^2 so that multiplying by x gives 2x^3

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 + ...) + .....

When -1 multiplies this quadratic term we get -1 times 2x^2 = -2x^2. But the x^2 term is -5x^2 so we are 'short' by -3x^2. That can only come from a linear term multiplied by x ... implies the linear term must be -3x

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x + ...) + .....

-1 multiplied by -3x gives +3x. We want -9x so -12x must come from somewhere. It must come from a number multiplied by x. So the number is -12

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x -12) + .....

-1 multiplied by -12 = + 12. We want + 18 so the remainder is +6

So: 2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x -12) + 6

Although I have not set this out as a division you should be able to check where you went wrong.

WHOOPS! That sin't any of the possible answers. Just checking what I have done ..... Back soon.

LATER EDIT I've multiplied out my expression and get 2x^3 - 5x^2 - 9x + 18. That's odd. I'm going to use some heavier algebraic hammers to crack this puzzle. Back soon.

EVEN LATER EDIT: I've checked all my working by another method and find the same result. Could these be the answers to another question entirely ? Or could the original cubic be different ?

Bob

Before I started I used the remainder theorem to find the remainder. I was expecting + 6 so that looks ok.

hi jose jose

You can achieve mathematical layout by using LaTex. There is a tutorial on this here:

http://www.mathisfunforum.com/viewtopic.php?id=4397

And there is a site called WolframAlpha that will compute some integrals for you. It's here:

But that won't help if you're trying to learn how to do them for yourself. If you post a problem on the forum someone will try to help, but don't expect we're just going to do your homework for you

If you cannot create the right LaTex try saying the question in words. We can translate those for you and post up what we think you're asking.

Bob

hi Hannibal lecter

Thanks for posting that. I wanted to check the wording ... in particular "measured in years before 2008". t might have been 2 yearly rather than one yearly.

So (b) First measure is 7.019. t value from table is 25 so 2008 - 25 = 1983. Your answer is correct.

Second measure is 6.957. t value is 125 so 2008 - 125 = 1883. Book answer is correct. If the question said 6.992 your answer would be correct, but both posts show 6.957.

Text books do sometimes have wrong answers. Mostly a typo, but in this case it looks like the question was changed (maybe to make it easier) but the writer forgot to alter the answer. Then you have to look further along for the next answer ... stopping you from just going along to the next column.

Hope that helps,

Bob

hi MathsIsFun

That's a wonderful moving diagram. I love it! Never seen that before.

Bob

hi Hannibal lecter

In the textbook, just before the table, there was an introduction to the question, explaining what the table shows, and what t and S stand for. This introduction is essential for anyone trying to do the question. Please copy that wording exactly as it is in the book.

Thanks,

Bob

hi ericnumberking

I don't have any direct experience of these so I've had a look at each. Study Pug has some free lessons so you could trial it yourself.

How did you arrive at this forum? Maybe you bypassed our very own teaching pages. The owner of the site, MathsIsFun, himself, has written a large number of maths help pages. They cover most topics up to calculus and each page is cross linked to related pages. There are many interactive lessons and all are written in an easy to follow style with clear diagrams and explanations. At the bottom of each page you will find a set of multi-choice questions to test your understanding. As an retired teacher I think these pages are excellent! And they're completely FREE. So before you commit money to any on-line course, why not try out MIF's own material.

Start here: http://www.mathsisfun.com/

And you can also ask for help from members of the forum.

Bob

I think it is valid. Compare with a standard induction proof for the sum of n integers.

I think the formula is sum= n(n+1)/2

step 1: when n = 1 , sum = 1x2/2 = 1

So the formula works for n = 1

step 2: assume it is correct for n = k

sum of k integers = k(k+1)/2

The next integer is k+1.

So the sum of k+1 integers is **sum of k integers + k+1 = k(k+1)/2 + k+1 = (k+1)(k/2 + 1) = (k+1)(k+2)/2**

This is the same formula with k replaced by k+1, so if the formula works for n=k then it works for n = k+1

The bit in bold characters is just algebra. So it is normal to include some algebraic steps in the proof.

Bob

Nothing hard. I must have explained it badly.

step (2): consider the expression k(k+1)(k+2) + 3(k+1)(k+2) = term 1 + term 2

The first term is divisible by 6 by (inductive) assumption.

The second term has 3 as a factor and 2 also as either k+1 or k+2 must be even.

So it also is divisible by 6.

Adding term 1 to term 2 must therefore be divisible by 6.

But the expression simplifies to (k+1)(k+2)(k+3) completing the induction step.

As you can see there is nothing original in this; it is just your proof written backwards.

Alg Num Theory: Hi. I agree but the title of the post seems to require a proof by induction. Perhaps it's in that chapter as an exercise in induction.

I once did some A level marking and one question told candidates to use a certain method. The chief examiner ordered us to give no marks if any other method was used!!!

Bob

hi Monox D. I-Fly

That looks good to me. You've done the two stages: (1) showed it works for a starting value; (2) showed that true for k leads to true for k+1. There will always be a bit of algebra at (2) and what you have done is correct and does prove stage (2).

As an alternative, you could start by saying let's consider the expression k(k+1)(k+2) + 3(k+1)(k+2). Show this is divisible exactly as you have done, and hence complete the stage (2) step. Essentially it's the same proof but backwards. I'm sure either way round is acceptable. Well done!

Bob

Today I visited your link page again and found someone calling themselves "Mohammad Matin" has copied my earlier post and submitted it as their entry. This seems to be doubly dishonest: (1) because it was not that person's work; and (2) it claimed to be a solution but definitely was not!.

It's a great shame that the work of members on this forum is being abused in this way.

Bob

hi Monox D. I-Fly

Sorry my post was too brief. I was in a hurry as I had to go out. Here's some more detail:

note:

Let A = 2x-5 and B = x+5 => A-B = x - 10 and A+B = 3x

The two trig. formulas that will help are

and

so the equation becomes:

I'll leave you to finish this.

Bob

I've received your message but I don't understand what you're wanting. I've visited the contest page and seen this:

"Contest Brief

Please find the problem attached to the file below."

But I cannot find any such attachment.

Bob

try using:

cos(A-B) + cos(A+B) = 2 cos(A) . cos(B)

A = 2x-5 and B = x+5

and a similar formula with a minus sign.

Bob

I think this can be done by calculating expected result for each route.

Assuming the plus and minus figures are equally likely:

the expected result for A is - (0.25 x 0.1 + 0.25 x 0.2 + 0.25 x 0.4 + 0.25 x 0.4) ** In your post you have 0.4 twice. If that is a typo then just substitute the correct value instead.** = - 0.25 x (0.1+0.2+**0.4**+0.4) = 0.25 x 1.1 = - 0.275

And for B = + 0.2 x 1 + 0.2 x 1.3 + 0.2 x 1.4 + 0.2 x 0.7 + 0.2 x 0.5 = 0.2 x (1+1.3+1.4+0.7+0.5) = 0.2 x 4.9 = + 0.98

So for A = Y + 1.5 - 0.275 = Y + 1.225

and for B = Y - 0.5 + 1.225 = Y + 0.725

Neither has an expected result within the limits. Am I misunderstanding something here?

Bob

formula for an expected result is

where the Ps are the probabilities and the Ws are the 'weights' for each possibility.

hi MrGiggles

Welcome to the forum.

I'm not sure I can help as I'm a bit confused about the problem. (no diagram seen)

Here's what I think you're asking:

There's a line connection between point X and point Y. Then the signal can either be sent to A, or to B. Either is acceptable if the signal stays within 0.5 of the original from X ( not higher nor lower than this limit)

There's no loss between X and Y (or it's irrelevant since we have to use this connection).

If A is chosen the signal will go up to value at Y + 1.5

If B is chosen the signal will drop to value at Y - 0.5

In addition transmission along YA and YB are subject to fluctuations as follows A route: minus 0.1,0.2,0.4,0.4 B route: plus 1,1.3,1.4,0.7,0.5

Is this correct so far?

Are these fluctuations the only ones possible? And is each equally likely?

Bob