Brain Teasers ? or 50 Years On........... ?

sjbabbey posted:

1. 13/28 √

2. 17/32 √

3. 1/2  hmmmm ?

Hi SJ, Nice try - yet again !

The first two answers agree with mine - well done !

On No 3, I followed the route of :- Prob of (at least 1 six) = 1 - Prob of (Less than 1 six)

          = 1 - Prob of (no sixes)

I found this approach to be more straight forward and my answer was...... (a little bit less than 1/2...........)

After klömdöngeling on my last answer as a result of not thinking things over properly, aided by a little thing that got lost in translation, here is my take on these three:

1. A probability of 0,46 for two identical items. My line of thought was 5/8 * 4/7 + 3/8 * 2/7

2. A probability of 0,53 for two identical items. My line of thought was 5/8 ^2 + 3/8 ^2

These are identical to SJs results.

3. A probability of 0,42 for at least one six. There are 6^3 (216) possible outcomes, 91 containing at least one six -> 91/216. I arrived at the 91 outcomes in a very roundabout way. Maybe someone has something elegant to show here.

I was about to say the same as SJ here as well. But 1/6 + 1/6 + 1/6 = 1/2 is not true. Why? Lets say you throw two coins. This line of thought would give a probability of 1 (sure thing) for getting heads (or tales) at least once. But only three of the possible four outcomes have heads at least once.

By the way, thanks Don for showing the right way to go at the bacteria  

Mulberry posted:

 

By the way, thanks Don for showing the right way to go at the bacteria  

You did really well....just one day out 

But the principle was right !

Mulberry posted:

After klömdöngeling on my last answer as a result of not thinking things over properly, aided by a little thing that got lost in translation, here is my take on these three:

1. A probability of 0,46 for two identical items. My line of thought was 5/8 * 4/7 + 3/8 * 2/7

2. A probability of 0,53 for two identical items. My line of thought was 5/8 ^2 + 3/8 ^2

These are identical to SJs results.

spot-on Mulberry with the probabilities..as was SJ

I worked in Örebro (which is in Sweden not Germany) for about 6 months in 1967 and in Germany off-and-on during 1980-1985, but can only guess what klömdöngeling means

Mulberry, you were spot-on with this one. Richard has a 91/216 chance of being the proud owner of a Statement (I hear rumours that it might have been upgraded to a Mu-so Qb )

My solution is posted above, but whether it is more elegant than your solution (or SJ's when he checks his arithmetic ) only you and he can decide.

Cheers

Don

A GCSE Level Pre-Power combination which Frank will deride...........

New-Naim, the latest recruit to the Forum, is trying to decide what pre-power amp combination to buy (I do so dislike that new term “combo”…….)

He is contemplating either a 282 or a 252 pre amp, both with a SupercapDR.
He is also contemplating a 200DR, a 250DR or a 300DR power amp

To date he has auditioned various combinations of these items at his friendly dealers and he has sought advice from the forum.

George has recommended the 200DR, Nigel the 250DR and Graham the 300DR (these are spoof names, obviously………..)

None of them were prepared to comment on the two pre-amps but instead suggested a more versatile 272 or a Statement S1 pre. Both of these suggestions have been rejected as have all others…….

New-Naim already has a competent source and a proper pair of speakers.

Having asked “Huge” questions, New-Naim has evaluated his preferences objectively but has yet to choose his combination.

Of the two pre-amps his probabilities of choosing are 252…2/3 and the 282…1/3

Of the three power amps the situation is a bit more complicated :-
If he chooses the 252 pre-amp he is inclined towards either the 300DR at 1/4 probability or the 250DR at 3/4.
If he chooses the 282 pre-amp he is inclined towards either the 200Dr at 1/5 probability or the 250DR at 4/5

What is the probability that his selected combination will include a 250DR ?

23/30 is my best guess.

BTW I can see where I went wrong on the previous question. Should have spotted the inverse probability solution as this was used on previous questions. Just acquired the 50% probability system today. Very enjoyable and will get even better !

sjbabbey posted:

23/30 is my best guess.

BTW I can see where I went wrong on the previous question. Should have spotted the inverse probability solution as this was used on previous questions.Just acquired the 50% probability system today. Very enjoyable and will get even better !

Well, you didn't go wrong this time sj, 23/30 is good..............

Congratulations on your new system, hope it gives you years of great musical pleasure !

BTW sj,

Feel free to explain how you arrived at 23/30 because not everybody will be familiar and one or two might wish to refresh their memories of schooldays gone by.

If you'd rather not, no probs, i'll draw up a tree diagram and post it in a day or so.

Probability of  252 and 250 = 2/3 * 3/4 = 6/12 = 1/2

Probability of 282 and 250 = 1/3 * 4/5 = 4/15

So probability of a 250 = 1/2 + 4/15

or 15/30 + 8/30 = 23/30

thanks for your kind words, Don.

Steve

I have a wonderful pair of JR149 speakers from way back when. You know the type, metal cylinder with a top and bottom plate held together by a concentric bolt and nut, putting the bolt into tension and the cylinder into compression. Makes the whole thing really stiff and rigid.

I work near Salisbury and my cousin is a dustman who works the Southampton Road commercial district collecting garbage, well…………he showed me these screwed up drawings with no Naim suggesting that a local hifi manufacturer was developing a new range ofhigh-quality wireless speaker called “Mu-Tube” based on tubes increasing in diameter throughout the range. The company works in decimetres (10cm = 1dm)

In order to standardise procurement and production (these things are designed in Salisbury, manufactured in China then “tuned-up” in Salisbury before dispatch to retailers…..sorry, I digress  ) the designer requires each tube to have a radius that is an integer. Production requires each tube to be placed on a horizontal production rack such that the tubes can be rolled along the rack from left to right.

Consider the movement along the rack from left to right to be the x-axis. The y-axis to be vertical, with y = 0 = the “table-top” of the production rack, and the axes of the tubes to lay along the z-axis.

For reasons that were unclear, production requires a pair of tubes, radius m and n to be placed on the production rack, in contact with each other and at integer centres along the x-axis (don’t ask, I don’t know why……. The screwed up papers were unclear)

Give the three smallest solutions for n when m = 1dm.

It all looked a bit convoluted,  so i've produced an illustrative picture to help out........

For reasons that were unclear, production requires a pair of tubes, radius m and n to be placed on the production rack,in contact with each other and at integer centres along the x-axis (don’t ask, I don’t know why……. The screwed up papers were unclear)

Give the three smallest solutions for n when m = 1dm. (m = diameter of the red tube and n = diameter of the yellow, or blue or green tubes - the green one isn't shown! )

Don

I get the 3 smallest diameters as:  4dm; 9dm; 16dm

Regards,

Peter

On the ball as usual Peter

I think I'll have to go back to the A-Level and First-Year university maths books........

Of 30 drivers interviewed, 9 had been involved in a car crash at some time. Of those that had been involved in a crash, 5 wore glasses.  The probability of wearing glasses given that an interviewed driver had not been involved in a car crash is 1/3.

1. What is the probability that one of the 30 drivers, chosen at random, wears glasses ?
2. What is the probability that a glasses wearer has been involved in a car crash ?

Errrr, yes, this one appeared in an A-Level book. I know, I know. Standards these days..................

Given the current UK interest in space due to Tim Peak on the ISS (btw, I have in the past flown with Tim, or rather he has flown with me !) I thought the following (simplified) mechanics question might stir one or two A-Level grey cells……….

A rocket with mass 1,400,000 kg is launched vertically upwards by engines providing a force of 34,000,000 N.

1. Assuming the only other force on the rocket is its weight, what is its acceleration.
2. The actual acceleration is 12m/s², what is the magnitude of the total resistive force R acting on the rocket
3. How long will it take to reach a height of 20km (to the nearest second)
4. At a height of 20km, the engines stop and the rocket continues freely under the action of gravity only (the resistive forces can be ignored). What maximum height will the rocket reach

Assume all the following (these are the main) simplifications:

g = 9.8 m/s² and constant
R is constant up to 20 km (then zero)
The force from the engines is constant
The mass of the rocket is constant

Don Atkinson posted:

Of 30 drivers interviewed, 9 had been involved in a car crash at some time. Of those that had been involved in a crash, 5 wore glasses.  The probability of wearing glasses given that an interviewed driver had not been involved in a car crash is 1/3.

1. What is the probability that one of the 30 drivers, chosen at random, wears glasses ?
2. What is the probability that a glasses wearer has been involved in a car crash ?

Errrr, yes, this one appeared in an A-Level book. I know, I know. Standards these days..................

When Frank (Fatcat) suggested above that I had been hi-jacked, I thought I had better up the ante and move on from GCSE level. This one came out of the AS sylabus, so not quite A-Level proper, and hence probably not worth bothering with..........

Anyway, although I found the first bit easy enough (5+7 out of 30), I had to draw a tree diagram to visualise the second part, and even then it took me a few moments to appreciate what I was looking for.

Don.

I wasn't complaining about the easy questions, by the way.

Since you've upped the anti, I'm struggling to even understand the question.  

OK I'll bite but not confident at all.

a. 14.486 ms∧2

b. 3,480,000N

c. 58 seconds

d. 31.755 Km 

a. Using f=ma a force of 34KN on a mass of 1,400,000 will produce an acceleration of 24.286 m/s/s. Deducting the deceleration due to the force of gravity gives a net acceleration of 14.486 m/s/s

b. Again f=ma. R = deceleration 2.486 m/s/s (14.486 - 12 m/s/s) x 1,400,000 Kg = 3,480,000N

c. Using s = ut + at.t/2 to solve for t when u = 0, a = 12 m/s/s and s = 20,000m

d. Using v2 = u2 + 2as (twice) to find the velocity at s = 20,000m when a = 12 m/s/s and u = 0. Then finding the additional height (s) with this starting velocity and terminal velocity v = 0 when a = -9.8m/s/s

fatcat posted:

Don.

I wasn't complaining about the easy questions, by the way.

Since you've upped the anti, I'm struggling to even understand the question.  

Could be my poor Use of English.............

sjbabbey posted:

OK I'll bite but not confident at all.

a. 14.486 ms∧2

b. 3,480,000N

c. 58 seconds

d. 31.755 Km 

a, b and c are all spot-on SJ so no need to lack any confidence !!!

I'll have to re-check d, (but i'm confident my answer is right )

Well, pretty confident .............. providing I wrote the right question !

Yup, I see that for answer d, I've found v2 as 480m/s but then squared it again to get an additional 11,755m above the original 20,000m. 

Instead it should be 0 = 480 - 2 x 9.8s

i.e. s = 480/19.6  = approx 24.5m above the original 20Km

My workings, more to show that I do try these out before posting............I'm not getting you guys to do my homework for me - honestly 

Don Atkinson posted:

1. What is the probability that a glasses wearer has been involved in a car crash ?

 

try a tree diagram, it might help...........