The h Bar

Sir Cumference

5:00 AM

I've seen scum who were joking about the Nice attacks

0celo7

You need to hang out in antisemitic circles more.

Sir Cumference

minutes after the attack

On a reddit chat

0celo7

Reddit has chats?

Sir Cumference

People make chats on other sites when big events happen, and tons of people join

"Boom boom! Look out kids, it's another bomb!"

That's what one person said

Minutes after it happened

Crap, it's 1:00am

Gotta go to sleep

'Night

0celo7

Night.

user116211

5:02 AM

@SirCumference I'm alive till 3.

user116211

@SirCumference o/

0celo7

I'm gone too.

cheerio

user116211

@0celo7 night, my new god.

user228700

@JohnRennie: Morning sir. Are u there?

John Rennie

@KaumudiHarikumar Morning

user218912

5:04 AM

can't sleep.

user228700

I've got a question regarding free body diagrams and constraints. Can you please help?

user116211

@IceLord do some heavy computations.

John Rennie

@KaumudiHarikumar I can try. What's the question?

user228700

Here:

user218912

@MAFIA36790 lol the last question in my problem set.

user218912

5:05 AM

have you compute the e-l equations for a huge lagrangian for a vector field.

user116211

@IceLord hmm, then... take ~~some~~ one sleeping pill.

user116211

@IceLord Nope.

user218912

i don't have one.

user116211

@IceLord I'm into calculus of variations now.

user218912

@MAFIA36790 just know how to derive the E-L equations

user218912

5:07 AM

that's all the calculus of variations you need.

user228700

user228700

I had figured this out before but I don't think I understand it correctly; I'd gotten it right by chance.

user116211

@IceLord Well, Lanczos introduces the E-L equations in $(211.1)$; so I'm not behind that.

user228700

Give me a few moments to explain my difficulty; I'll try to figure it out again.

John Rennie

That looks quite a fun question, though it seems a hard question for someone not at university yet.

user116211

5:11 AM

Of course for $n$ degrees of freedom.

user228700

Oh crap, didn't even say what the question is!

user218912

@JohnRennie you're weird if you think that's fun xP

user116211

@IceLord He's a nerd, dude ;))

user228700

"In the figure shown, blocks A and B move with velocities v1 and v2 along the horizontal direction. Find the ratio v1/v2 if the string is inextensible and incompressible"

John Rennie

@IceLord I think it is quite a neat question because there are various ways to approach it.

user228700

5:12 AM

@MAFIA36790 Oh my God, like literally everyone else here isn't!

user218912

@KaumudiHarikumar @0celo7 isn't.

John Rennie

@IceLord he says he isn't ...

user228700

@IceLord I guess it depends on your definition of "nerd". Don't have the time to discuss that now :P @JohnRennie: I'm trying to figure it out again, please hold on.

John Rennie

@KaumudiHarikumar OK ...

Xasel

hmm..first we will account for forces acting on the body

Tension

Tcos theta is acting horizontaly

DanielSank

5:15 AM

@KaumudiHarikumar That has always bothered me.

user116211

Okay, @JohnRennie, from yesterday, I'm trying to conceive Qmech's answer....

DanielSank

I have no idea why so many people use "doubt" in that way.

user116211

I know Euclidean space is an affine space.

Xasel

and T sin theta + N_1 - mg =0 (holding it to froun

user218912

@MAFIA36790 what?

Xasel

5:16 AM

@KaumudiHarikumar:is the masses are same?

user116211

He says that by introducing constraints, it becomes an embedded n-dimensional submanifold which is not Euclidean.

John Rennie

@Xasel the masses don't matter

DanielSank

@0celo7 for whatever it's worth, I find your tone in chat lately somewhat unpleasant.

user116211

> The submanifold inherits a Riemannian metric by pullback from the ambient Euclidean space.

user116211

What is pullback?

user228700

5:18 AM

Yeah, so I think I've got it.

user116211

@IceLord what?

Xasel

@JohnRennie: correct me where I'm going wrong but ain't : T sin theta + N_1 - mg =0 implies that mass matter

@KaumudiHarikumar:Care to share the soln

user228700

@DanielSank We people from India do. Safe flight?

DanielSank

@MAFIA36790 Did you look it up?

user218912

@MAFIA36790 euclidean space is not an affine space period. it's an affine space with some extra structures.

John Rennie

5:19 AM

@MAFIA36790 I don't think that's a technical term, and I don't think it has a special meaning.

user116211

@DanielSank yeh, he linked the wiki article.

DanielSank

@KaumudiHarikumar I see that now from the post you linked. Very interesting. This explains a lot of what I thought was "very weird wording". Yes the flight was good.

user116211

@IceLord Of course, angles, the metric.

John Rennie

If you take a manifold with a metric then any submanifold has a metric that is related to the parent.

user228700

Yep, so I made use of the fact that every part of the string moves with the same velocity.

user218912

5:20 AM

gonna sleep now 4realzzz

user218912

bye

Xasel

so what's the answer (is it = 1)

user116211

@IceLord o/

John Rennie

@MAFIA36790 I think this called an induced metric

@KaumudiHarikumar that's the way I would do it.

user228700

Let's look at the part connected to the block A first. (I need to be corrected if this reasoning is wrong @JohnRennie. That's why I was wondering if u could.) The horizontal component of the velocity of the string is the same as the velocity of the block A. So if $u$ be the velocity of the string moving up, we get ucos(theta1)=v1

Xasel

5:24 AM

what's the answer BT?

user228700

At the other end, the string is moving in the opposite direction with the same speed. Resolving the velocity vector into components again and equation the horizontal component to v2, we get ucos(theta2)=v2

Xasel

well I get it but something is bugging me

user228700

Using these two equations, we get v2/v1=cos(theta1)/cos(theta2). This is the answer given in my answer key as well but @JohnRennie, is my reasoning correct?

Xasel

Ain't after bloack A cross the mid-point of pulley there will be (de)acceleration?

user228700

It was a little difficult for me to wind my head around the fact that the horizontal component will be the same as the velocity of the block...

user228700

5:29 AM

@Xasel Huh? No, we're asked to look into the situation for this particular moment alone.

John Rennie

What you call the velocity of the string is really the rate of change of the distance from A to the pivot.

user228700

@JohnRennie Yes, OK...

John Rennie

If we call this distance $l$ then the rate of change is $dl/dt$, and because the string has a fixed length this has to be the same for both sides of the string.

user228700

@JohnRennie Yup...

Xasel

but ain't until and until Block a arrivess(or crosses) the midpoint of puley

There won't be any change

John Rennie

5:31 AM

Consider the distance from A to the point immediately under the pivot. Call this $l_A$.

Then $l_A = l \cos\theta_1$

The speed of A is just $dl_A/dt$, so we can take the equation above and differentiate both side wrt time to get:

user228700

Wait, what do you mean by immediately under? Which point is that?

John Rennie

@KaumudiHarikumar I'll draw a diagram.Hang on ...

user228700

@Xasel I'm sorry but I don't understand what you're trying to say...

user228700

@JohnRennie Okay!

user228700

Wait, do you mean the point under the pulley, on the same horizontal level as the block? Your equations make sense if that's the point u were talking about...

Xasel

5:36 AM

But there is no change in length of string untilt he block a Crosses the pivot point

it startts to happen after that and the we have to account for aftermath of retardation

John Rennie

So obviously $l_A = l \cos\theta_1$

Xasel

ok got it

user228700

Yes...

Xasel

but ain't we have to account for aftermath of retardation after a time intervall

John Rennie

And the velocity of A is related to $l_A$ by: $$v_1 = \frac{dl_A}{dt}$$

user228700

5:40 AM

@JohnRennie OK...

user116211

Yes, @JohnRennie:

user116211

> [...] induced metric is the metric tensor defined on a submanifold which is calculated from the metric tensor on a larger manifold into which the submanifold is embedded.

John Rennie

And (I think this is true): $$\frac{dl_A}{dt} = \frac{d}{dt}(l\cos\theta_1) = \frac{dl}{dt}\cos\theta_1$$

So in your notation $v_1 = u\cos\theta_1$

user228700

@JohnRennie Assuming that for that instant, theta is a constant, yes?

John Rennie

@KaumudiHarikumar That's the only thing I'd have to thing about. Should we differentiate $l\cos\theta_1$ using the product rule? I'd have to go away and think about it.

user228700

5:44 AM

(I know that sometimes, I state what is clearly very obvious already but I just want to make sure...)

John Rennie

The product rule would give us: $$ \frac{d}{dt}(l\cos\theta_1) = l\frac{d}{dt}(\cos\theta_1) + \frac{dl}{dt}\cos\theta_1$$

user228700

@JohnRennie Yeah, no, this would be correct. We've only been asked to calculate v2/v1 for the given instant described by the figure so I think that assuming theta to be constant is fair enough.

user116211

@KaumudiHarikumar Just because it's about the instant, that doesn't mean you should take $\theta$ constant. You have to use the product rule.

user228700

@MAFIA36790 Why not? We only care about the instant and in the instant, theta remains the same. But I can see how this sort of reasoning might land me in trouble; for this instant, l is also constant so by this reasoning, the velocity of the string would come out to be zero. OK, great.

user228700

@JohnRennie: Thoughts?

user116211

5:50 AM

@KaumudiHarikumar I'm always right ;P

user228700

@MAFIA36790 Sure :P

user228700

@JohnRennie: Did u go away to think about it? :P

John Rennie

@KaumudiHarikumar I'm still thinking ...

user228700

@JohnRennie OK sure.

John Rennie

OK go back to my drawing, and call the hieght of the vertical line $h$

So by Pythagoras $l_A^2 + h^2 = l^2$

user228700

5:54 AM

OK...

Xasel

@John Rennie:SHouldn'e we use this:awwapp.com

awwapp.com/b/uaagipfol

John Rennie

Now differentiate both sides wrt time. $h$ is a constant so $\frac{d}{dt}(h^2) = 0$

user228700

Right.

John Rennie

$\frac{d}{dt}(l^2) = 2l \frac{dl}{dt}$

And likewise $\frac{d}{dt}(l_A^2) = 2l \frac{dl_A}{dt}$

Are you happy with this step? This is a trick you may not have seen before ...

Xasel

Hey is math is properly rendering with you guys

^It only shows me latez

John Rennie

5:58 AM

@Xasel you need an addon to render Latex in the chat. Google chatjax.

user228700

@JohnRennie In the second one, is it really $l$ on the right hand side and not $l_a$?

John Rennie

Oops, yes that's a typo. It should be $$\frac{d}{dt}(l_A^2) = 2l_A \frac{dl_A}{dt}$$

user228700

OK :-) Yes, I can see what ur next step would be.

user228700

So we haven't even used theta while differentiating! Nice!

John Rennie

Yes, we're going to get: $$2l_A \frac{dl_A}{dt} = 2l \frac{dl}{dt}$$

Hang on ...

user228700

6:02 AM

Yes, and then we know that d/dt(l_a)=v1 and d/dt(l)=u. We also substitute for l_a as lcos(theta1) and viola, we get u=v1cos(theta)

John Rennie

That's going to give us: $$v_A = u\frac{l}{l_A} = \frac{u}{\cos\theta_1} $$

Oh, OK, that's right.

user228700

This is a handy little trick that you've used! I've used it before but it didn't occur to me to use it now...

John Rennie

@MAFIA36790: you still here? Sorry, I got distracted with K's problem and didn't reply to your comment.

Xasel

@KaumudiHarikumar:he is the gandalf here he no which concoction of spell to use :p

user228700

Okay, so my basic doubt before all this was relating the string's velocity and the objects's.

user228700

6:08 AM

@Xasel This is the second time he's been referred to as Gandalf :-)

Xasel

@KaumudiHarikumar:Actually he is,he time travelled back to '30s and inspired Tolkien to write LOTR

user228700

I wasn't able to understand how to make sense of the fact that the horizontal velocity of the string would be the same as the velocity if the block. And I'm required to use this concept several times, in different problems.

John Rennie

@KaumudiHarikumar: NBBBB. You started with:

46 mins ago, by Kaumudi Harikumar

Let's look at the part connected to the block A first. (I need to be corrected if this reasoning is wrong @JohnRennie. That's why I was wondering if u could.) The horizontal component of the velocity of the string is the same as the velocity of the block A. So if $u$ be the velocity of the string moving up, we get ucos(theta1)=v1

user228700

I think I've even posted a question about this.

John Rennie

i.e. $u\cos(\theta_1)=v_1$

user228700

6:10 AM

:o What's NBBBB?!

John Rennie

And we've just proved that $\frac{u}{\cos(\theta_1)}=v_1$

user228700

Yeah, I understand the logic now, but it was difficult for me to before.

John Rennie

NB is nota bene, Latin for note well, so NBBBB is note very, very, very well :-)

user228700

Basically, the string and the block don't move independent of each other, which is where all this comes from, yes?

Xasel

@MAFIA36790:that day actuallyy I meant after passing the 51st or 57th page..do tell me that whether the book is suitable for beginner(ok this seems superstition butin my opinion books tarts to open up after that page..and show their true colours)))

@KaumudiHarikumar: Bounty for those

who resolve cos theta ,accouunting for change in theta over time

user228700

6:13 AM

@JohnRennie OK, noted :-)

user116211

@Xasel Most of the new books start to define vector space after 250-300; it starts right from page no.28. Now you decide.

John Rennie

@MAFIA36790 ah you're back. Still want to talk about that manifold thing?

user228700

BTW, did u know that there is a school in Canada named "John Rennie school"?

user228700

Same spelling too!

John Rennie

There are quite a few famous John Rennie's. Sadly I am not one of them.

user116211

6:15 AM

@JohnRennie sure; but I'll take some time to read some other stuff... I'll ping you.

John Rennie

@MAFIA36790 I can explain it really quickly and easily ...

user228700

@JohnRennie Oh, I see. I'd argue that u're pretty famous for ur knowledge on this website and that's more than what a lot of people get, and I'd argue that it's more valuable even :-)

Xasel

@MAFIA36790: hmm...well I only know basic stuff about vector space taught n HS not above so I can't decide..that why I am asking you opinio sir

user116211

@JohnRennie That would be great! But just a few mins; I'll ping you.

John Rennie

@MAFIA36790 you have 14 minutes then I have to start work

user116211

6:16 AM

@JohnRennie Well, then go on ;))

John Rennie

@MAFIA36790 Suppose you have a particle moving in 3D Euclidean space. The metric is obviously flat.

user116211

yep.

John Rennie

Now add a constraint that the particle must remaina fixed distance froma specified point, so the particle now moves on the surface of a sphere.

user116211

okay...

John Rennie

The metric of a sphere is not flat

user116211

6:18 AM

sure.

John Rennie

So there's an example of how imposing a constraint selects a submanifold that is not Euclidean.

user116211

okay!

user116211

@JohnRennie, As I know, for the line-element to be Euclidean, the terms like $\mathrm dx_i\mathrm dx_j$ must vanish. Is it true for all coordinate-system? As for the first case, where there was no constraint the line-element was in Riemannian form with those terms; but the geometry was still Euclidean. So, does that mean even with the presence of those terms, the geometry can be Euclidean?

John Rennie

Remember that you can choose any coordinate system you want. The presence of cross terms doesn't necessarily mean the space isn't flat, it could just mean you've chosen screwy coordinates.

user116211

@JohnRennie You made the things quite clear, thanks!!

John Rennie

6:24 AM

The only way to tell for certain is to calculate the Riemann curvature tensor. If space is flat this is zero in all coordinate systems.

user116211

@JohnRennie yeh, yeh, sure.

Xasel

what are screwy coordinates :P

user116211

Oh, @JohnRennie, one more thing, Qmech also left a comment behind all on a sudden last night:

John Rennie

@Xasel helical coordinates!

user116211

Comment to the post (v1): Consider to double check for $x\leftrightarrow q$ typos, and $3N \leftrightarrow n$ summation limit typos. — Qmechanic ♦ 16 hours ago

user116211

6:26 AM

I didn't get what he wanted to point out exactly.

John Rennie

I don't get it either. Sorry. It's going to take me more thatn the three minutes I have left before I have to start work!

Xasel

@MAFIA36790:Care to reply about my query ,sir?

user116211

@Xasel Hey, don't call me sir!

Xasel

ok "Respected Sir"

user116211

@Xasel oh man!! I'm just another comrade. I'm an undergrad; know very little things to be addressed as sir ;/

user116211

6:30 AM

@Xasel BTW, which query?

Xasel

knowledge is a relative scale so as compared to me you are : High Commandant COmrade

Repeating the above text: hmm...well I only know basic stuff about vector space taught n HS not above so I can't decide..that why I am asking you opinio sir

user116211

@Xasel That sounds good ;)

user116211

@Xasel Yes, sure; you can read the book; it explains from the base level. You don't need to know what vector space is to conceive the contents of the book. As I said, it is self-contained; so you can easily read it. Everything is developed there.

Xasel

@MAFIA36790:thankyou then..willb uy it after exams for killing free-time(BTW is this exercise based book or noly theoretical concepts with little problems here and there)

user116211

@Xasel yes it has its own exercise; but not much.

Xasel

6:35 AM

@MAFIA36790:Are you using other books too?

user116211

@Xasel on Linear Algebra?

Xasel

yeah

user116211

@JohnRennie Will wait....

Xasel

@MAFIA36790:I wonder maths book be so interesting if they include exercises for implementing the module in book...

user116211

@Xasel Well, I don't have an actual book; but pdf of David C Lay and Axler. the former is pretty applied and contains many examples. The second is quite famous for its ditching on determinants. and some books on Classic algebra do contain some contents of Linear Algebra too.

Xasel

6:38 AM

what's the difference between classic algebra and linear algebra

user116211

@Xasel Well Classic algebra means the polynomials, and roots and all that. Those books have an extended chapter on linear algebra.

Xasel

oh...thanx for enlightening me on this

Danu

9:11 AM

disclose.tv/news/…

lololol

user228700

9:27 AM

Hi :-) Is anybody here? (Using the mob. chat-can't see)

user116211

@Danu What? Oh man ;P

user3183724

10:57 AM

Bit of a technical question for those who know about superconducting circuits. In what regime of E_J/E_C (josephson energy over charging energy) do cooper pair box qubits (or split cooper pair box qubits) typically operate? I know transmons are in the region of tens to hundreds and that CPBs are lower, but I am not sure what typical values are for those guys

Danu

11:20 AM

Hey @yuggib, long time no see

I think this is a really crucial question for potential moderators (taken from Grace Note's "standard questions"):

0

A: 2016 Moderator Election Q&A - Question Collection

I've taken this question from the "standard questions" answer by Grace Note: It is one of the optional questions. I'm putting it in a separate answer because it will otherwise probably not be selected, while I think it's a very good question. In what way do you feel that being a moderator wil...

Physics Meta

0

Q: What changes will the upcoming election bring?

So, apparently, we're 'scheduled' to have a moderator election later this month. What changes, exactly, will this election entail? Are any of the existing moderators stepping down? How many new moderators will be elected? Is the number of moderators changing? If so, does this mean that the curren...

Danu

Also lol, at least this guy is honest:

The point is that: it is one of our possible exam question and i have absolutely no idea on how to solve it , and i also did some research but didnt find anything reasonable . — Danijel Keno 20 hours ago

John Rennie

@Danu ironically I think being a mod would make me less effective. For example it would make me reluctant to vote to close, except possibly as the fifth vote. Except for the most blatant abuses I suspect I would review far fewer questions.

user228700

@JohnRennie Hey! Are u relatively free at the moment?

yuggib

11:36 AM

@Danu Hey, how are you doing?

Danu

@JohnRennie It definitely made me think: Hmm, what good sides are there even?!

yuggib

I'll be there more now that summer is finished (up to a set of measure zero of days)

Danu

@JohnRennie It certainly makes your reviewing go down (basically to 0)---I've noticed this as a moderator of HSM

@yuggib Nice

@yuggib Pretty good

Working on a complex geometry book

As in reading, of course :P

yuggib

you're now a pure bred mathematician, I see... ;-P

Danu

Hope so :) I've very nervous (already!) about finding a PhD spot :\

It seems to come down to having a supervisor who either keeps you around or hooks you up with someone.

I hope my supervisor will do either :P

yuggib

11:45 AM

I am sure you will find a place in the end

you can also look for somebody that does things you like, and contact him directly

in any case I have the impression that obtaining funding to do a phd in germany is not difficult

Danu

Yeah, but I've seen from my friends around here that that has a... <10% chance of working out.

Nobody has funding :P

yuggib

yeah you have funding through courses or GRK

Danu

GRK? courses?

yuggib

you get paid for doing the teaching assistant in courses

Danu

Sure, but that's about 400 per month.

You need like 3x that (which would amount to 25+ hours work per week) to have a reasonable salary

The whole point of the PhD is getting paid to do research :P

yuggib

11:50 AM

I have heard people working much less and get a reasonable salary

in germany

anyways, the other solution is to find some professor with a grant (e.g. Graduiertenkollegs or european grants) that pays PhD students with those funds

finally, you can think of changing country ;-P

user116211

@yuggib o/

yuggib

@MAFIA36790 \o

John Rennie

12:32 PM

@KaumudiHarikumar: I'm around for a few minutes if you're still here ...

user228700

I'm still here...

user228700

How many minutes is "a few minutes"? (:P)

John Rennie

Well ask your question and we'll see how long it takes :-)

user228700

OK :-) I dunno if you'll be ready to answer though, 'cause it's not physics, it's math.

John Rennie

OK, well give it a try ...

user228700

12:36 PM

Albeit, high school math so you're gonna be able to answer and all.

user228700

It's to do with probability.

John Rennie

OK

user228700

You know Baye's theorem, no?

John Rennie

I know of it though I don't know the details.

Bayes' theorem not Baye's theorem

user228700

Oh! No problem, I'll ask anyway. Pretty sure u'll be able to answer regardless.

user228700

12:39 PM

@JohnRennie Oh, OK. Noted. Thanks :-)

user228700

I'll type the question so I don't have to take a general case and try to explain very sloppily.

user228700

(Also, I tried asking at the MSE chat but nobody has responded in hours :-()

user228700

Okay, so say there's this guy, Tom. He can go to his office by three means; scooter, car, bus and train.

John Rennie

Just post the link to the MSE question

user228700

No, I didn't ask the whole question there. I just asked if anybody could help with a probability question.

John Rennie

12:43 PM

OK, go ahead, sorry for interrupting ...

user228700

Anyway. So Tom. The probabilities of him choosing any one of these four means is given.

user228700

NP.

John Rennie

oops, sorry, go ahead

user228700

Crap, I wrote three but mentioned four, sorry.

user228700

Okay, so these probabilities are given and they should add up to 1, no? 'Cause those are the only possibilities.

John Rennie

12:45 PM

Yes, the four probabilities should add to one

user228700

Okay, then they've given me the probabilities for him being late to work if he chooses one of these 4 and this is different for the four different means BUT they don't add up to one :/ I'm not able to make sense of this.

John Rennie

They don't have to add up to one.

user228700

I'm not even sure why I think they should add up to 1 actually :/

user228700

Why don't they have to?

John Rennie

Probabilities add up to one when they are the only possible outcomes.

user228700

12:48 PM

@JohnRennie Yup.

John Rennie

So P(being late) + P(not late) = 1 because either you are late or not late. You have to be one or the other.

user228700

So considering the event that he's late, he can only be late by one of those 4 means, right?

John Rennie

So suppose Tom takes the scooter then P(scotter,being late) + P(scooter,not late) = 1

user228700

Oh, you mean that the total probability of him being late doesn't have to add up to 1..?

John Rennie

Because when he is riding the scooter he can only either be late or on time.

There are eight possible outcomes: scooter/late, scooter/not late, car/late, car/not late, and so on.

user228700

12:51 PM

So P(Choose scooter).P(Late)+P(Choose scooter).P(On time)=1?

John Rennie

And the probabilities of those eight outcomes have to add up to one because they are the only options.

user228700

Wait, that doesn't make sense.

user228700

Yeah, OK, add that term with all the other terms and then we get 1.

John Rennie

Let P(scooter,being late) be the probability of being late if Tom takes the scooter.

user228700

'Cause the only thing that has to happen is that he gets to office, late or not...right?

John Rennie

12:54 PM

Yes, the probabilities for all the possible ways of arriving at the office have to sum to one because Tom always arrives at the office.

user228700

So I look at the event that happens no matter what and add all the ways in which that can happen to get 1, yes?

John Rennie

As I said, let's take P(scooter,late) as the probability of being late if Tom takes the scooter. So P(scooter,not late) is the probability of being on time if Tom takes the scooter.

if Tom takes the scooter he always arrives at the office, so P(scooter,late) + P(scooter,not late) = 1.

ACuriousMind

@yuggib Actually, as far as I know, explicit teaching assistants are only bachelor's and master's students. Everyone else doing that is doing that as part of their salaried position as doctoral student/postdoc, and they don't get paid extra for that teaching

user228700

@JohnRennie Ohh.

John Rennie

The probability that Tom both takes the scooter and is late is then P(scooter) \times P(scooter,late).

user228700

12:57 PM

Yup, OK.

John Rennie

And the probability that Tom both takes the scooter and is on time is then P(scooter) \times P(scooter,not late).

user228700

So pick one way and look at all the sub-ways (God, am I being sloppy. Forgive me please.) and then add that to get 1.

John Rennie

Where we came in is you asked is:
P(scooter,late) + P(car,late) + P(bus,late) + P(train,late) = 1

user228700

Never mind if u didn't understand my truly horrible wording right there. I think I've got it and that's enough :P

user228700

@JohnRennie Yup. That was my original doubt.

John Rennie

12:59 PM

Do you see now why the answer is no?

user228700

Yes, I do :-) Thank you!!

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