3 Quickies

[rocdocmac]

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 (6^th) 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 _

 

[bonanova]

Ten minutes for the bunch ...

Quickie 1

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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

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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.

Quickie 3

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

 

[rocdocmac]

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(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 February 2, 2018 by rocdocmac
Expanded response

[bonanova]

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1. Changing to probability of 5/12
2. Adding 12.5 mm to previous answer
3. No clue

[rocdocmac]

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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!

 

 

[flamebirde]

#3:

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Out of pure optimism, I'll say 1.

#1:

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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%.

 

[Molly Mae]

#3

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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.

[Thalia]

#1

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1/4?

[plainglazed]

1.

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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.

[rocdocmac]

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

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No value for #2 posted yet!

 

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#3 goes to @Molly Mae!

Yes, it is 6 1 3 1 4 5

[bonanova]

  On 2/2/2018 at 7:37 AM, rocdocmac said:

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(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.

 

Expand  

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

1. WWWW  6 ways to pick (WW) each with p(W_next) = 1 -- 6x1=6 desired outcomes
2. WWWB 3 ways to pick (WW) each with p(W_next) = .5 -- 3x1=1.5   "
3. WWBW 3 ways to pick (WW) each with p(W_next) = .5 -- 3x1=1.5   "
4. WBWW 3 ways to pick (WW) each with p(W_next) = .5 -- 3x1=1.5   "
5. WWBB 1 way to pick (WW) each with p(W_next) = 0 -- 1x0=0        "
6. WBWB 1 way to pick (WW) each with p(W_next) = 0 -- 1x0=0        "
7. WBBW 1 way to pick (WW) each with p(W_next) = 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(W_next) = 10.5/18 = 21/36

 

 

[rocdocmac]

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#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 February 3, 2018 by rocdocmac
Explanation refined

[flamebirde]

I hate to bring this one back up again, but what's the rationale behind #3?

[harey]

  On 2/1/2018 at 8:56 PM, rocdocmac said:

1 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?

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I do not get what the 12.5 mm are there for. Do you mean after the begin of the first song? I just ignore it and assume the first r=290 mm.

The usual name for these records in French is "33 tours" (to distinguish it from the previous standard of 78 rpm). (In fact, the speed is 33 1/3 rpm.)

In 18 minutes, the disk will make 600 rotations.

197 * 600 =~ 120,000 -> without any doubt, the last r=17 cm, average r=23 cm.

23 [cm] * 600 * 3.14 = 433 [meters].

I recall these LPs, the last r=17 cm seems a little bit high to me. They had so about 20 minutes playtime on each side, but I would guess the inner r nearer to 7 cm then.

 

Edited March 23, 2018 by harey

[Molly Mae]

  On 3/23/2018 at 3:43 AM, flamebirde said:

I hate to bring this one back up again, but what's the rationale behind #3?

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It's just a letter substitution.  1 = A, 2 = B, 3 = C, etc.

[flamebirde]

  On 3/23/2018 at 2:04 PM, Molly Mae said:

It's just a letter substitution.  1 = A, 2 = B, 3 = C, etc.

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gotcha, thanks.

[rocdocmac]

  On 3/23/2018 at 10:56 AM, harey said:

I do not get what the 12.5 mm are there for. Do you mean after the begin of the first song?

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The needle goes down 12.5 mm to the very beginning of the first song, i.e. before the first song actually starts.

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Pay attention to both Bonanova's remarks on 2 February 2018. Otherwise you'll go off track!

 

Edited March 26, 2018 by rocdocmac

[rocdocmac]

@harey...

Êtes-vous Franҫais?

[harey]

@rocdocmac

Pas du tout. Mais c'est ma langue maternelle, je sais même écrire.

 

[rocdocmac]

Merci beaucoup Harey et bonne nuit ...