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rocdocmac

3 Quickies

Question

10 minutes each!

1.      A bag holds four counters.  One of them is white.  Each of the others is either black or white at equal chance.  One randomly draws out two counters, and discover they are both white. If one then randomly draws a third counter, what is the chance that it is white?

 

2.      I’m about to play the music songs on Side 1 of a standard LP (diameter = 292 cm) that contains six tracks. The recorded surface area is 487.6 dm2 (including the “gaps” between each song) and the average distance between each annulus (i.e. distance between each “groove” along the LP radius) is 197 µm. The total time that it will take to play this side right from the beginning of the first song to exactly the end of the last (6th) song is 18:04 minutes. If the “needle” (stylus) is poised 12.5 mm directly above the beginning of the first song and 2 mm away from the edge of the record, how far does the “needle” travel to the point where the last song just ended?

 

3.      What comes next?

6 1 3 1 4 _

 

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11 answers to this question

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Ten minutes for the bunch ...

Quickie 1

Spoiler

There are four cases that fit the description of one W and three that are W/B with equal likelihood:

  1. W W W W
  2. W W W B
  3. W B B W
  4. W B B B

Case 4 is eliminated because 2 W were drawn, leaving three cases with probabilities of 1, .5 and 0 respectively of drawing another W.

The probability of drawing another W is thus 0.5.

If it weren't a "quickie." I might discuss the matter with Bayes and say that if at random the first two counters drawn were W, then it could be argued that case 1 is more likely, and Case 3 is less likely, than case 2. That would weight the desired probability to a higher value. But that would take me, at least, longer than 10 minutes. So I'll stick with 0.5. Also, Bayes confuses the crap out of me.

Quickie 2

Spoiler

OK so I don't have the determination to calculate it, but the needle moves the distance between the maximum and minimum radii of the recorded surface. There's enough information to calculate it, but ... someone else can do it. B))

Quickie 3

Spoiler

What I see here is 31416 in some order. What's missing, of course is the decimal point.

 

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Spoiler

(1) There's more to it than meets the eye! Forget the 10 minutes .... take your time! Think of W1, W2, W3, W4, B1, B2, etc.

(2) Does anyone wish to do some calculations? Again, don't worry about the 10 minutes.

(3) The decimal point is not missing ... there isn't one!

 

Edited by rocdocmac
Expanded response

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Spoiler

1. How did you arrive at 5/12?

2. A-ha!

3. If no one gets this, it won't be a 10 minute quickie anymore! Actually, this is a 5 second blitzie!

 

 

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#3:

Spoiler

Out of pure optimism, I'll say 1.

#1:

Spoiler

Well there's a 3/4*2/3=1/2 chance that I didn't draw the white disk the first go around.

In the scenario that i did draw the guaranteed white counter, then the chances of me drawing another white one are 1/2.

In the scenario that I didn't draw the guaranteed white counter, there's a 3/4 chance that I draw a white counter (1/2 to draw the guaranteed one, 1/2 to draw the coin flip, and for that coin flip 1/2 chance that it's white).

So, I would say that there is a 5/8 chance of drawing a white counter.

At least to me, it wouldn't make sense for the chances of drawing a white counter to be less than 50%.

 

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#3

Looking behind the facade, I'm going to have to say 5.  Once I realised it wasn't a sequence, it was the only thing that made sense.

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1.

Spoiler

Replace the three fifty/fifty counters with one black and one white then add another white counter to keep the odds of drawing the original sole white counter at 25%. So now starting with 5 white and 3 black counters I think gives the same odds as the original scenario.  And after picking two white, there are three of each left so 50%?

EDIT:  yeah, as I continue to wake up here, am less convinced of my logic above.

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Spoiler

#1 still wide open. Try making use of conditional probability.

Spoiler

No value for #2 posted yet!

 

Spoiler

#3 goes to @Molly Mae!

Yes, it is 6 1 3 1 4 5

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On 2/2/2018 at 2:37 AM, rocdocmac said:
  Hide contents

(1) There's more to it than meets the eye! Forget the 10 minutes .... take your time! Think of W1, W2, W3, W4, B1, B2, etc.

 

Spoiler

 

There are eight equally likely sets of marbles in the bag:

  1. WWWW  6 ways to pick (WW) each with p(Wnext) = 1 -- 6x1=6 desired outcomes
  2. WWWB 3 ways to pick (WW) each with p(Wnext) = .5 -- 3x1=1.5   "
  3. WWBW 3 ways to pick (WW) each with p(Wnext) = .5 -- 3x1=1.5   "
  4. WBWW 3 ways to pick (WW) each with p(Wnext) = .5 -- 3x1=1.5   "
  5. WWBB 1 way to pick (WW) each with p(Wnext) = 0 -- 1x0=0        "
  6. WBWB 1 way to pick (WW) each with p(Wnext) = 0 -- 1x0=0        "
  7. WBBW 1 way to pick (WW) each with p(Wnext) = 0 -- 1x0=0        "
  8. WBBB 0 ways to pick (WW)

Of the 18 ways (WW) could have been picked, there are 10.5 desired outcomes.

p(Wnext) = 10.5/18 = 21/36

 

 

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Spoiler

#1 goes to moderator @bonanova!

7/12 = 0.58333 ... (whether marbles/counters/balls/whatever were put in the bag).

 

Another way to calculate the answer using conditional probability (Thanks to Christian B):

The probabilities that we have (1,2,3,4) white counters in the bag are

(1/8,3/8,3/8,1/8)

Given that we have (1,2,3,4) white counters in the bag, the probabilities that we draw two white counters are

(0,1/6,1/2,1)

The overall probability P(2W) to draw two white counters is given by

P(2W) = 1/8*0+3/8*1/6+3/8*1/2+1/8*1 = 3/8

The probability that all three counters are white results in

P(3W) = 1/8*0+3/8*0+3/8*1/4+1/8*1 = 7/32

Thus, the conditional probability P(3W|2W) is given by

P(3W|2W) = P(3W∧2W)/P(2W) = P(3W)/P(2W) = 7/12 = 0.5833

 

 

 

 

 

Edited by rocdocmac
Explanation refined

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