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unJon
unJon
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December 2nd, 2020 at 11:02:45 AM permalink
Quote: CrystalMath

Quote: Gialmere


How many different sets of five primes can you form this way?




Each card must contain an odd and an even number, with the even preceding the odd.
05 is the only prime you can create with the number 5, so we only need to look at the ways to place the numbers 2,4,6,8 with 1,3,7,9.

We have the following possible primes:
41, 61
23, 43, 83
47, 67
29, 89

Because 41/61 choice will determine what number we use for 47/67, then we can never use the number 43.

This gets us down to:
41, 61
23, 83
47, 67
29, 89

So, it looks to me like there are only 4 choices:
05, 41, 67, 23, 89
05, 41, 67, 83, 29
05, 61, 47, 23, 89
05, 61, 47, 83, 29



Your first assumption is incorrect:

You can make a card with 02 which would then let you make a 53 card or a 59 card.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
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Gialmere
December 2nd, 2020 at 12:37:14 PM permalink
Quote: Gialmere

How many different sets of five primes can you form this way?



0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).
The only primes < 100 that contain a 5 are 5, 53, and 59
4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2
If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29
The two-digit primes that begin with 4 are 41, 43, and 47
The two-digit primes that begin with 6 are 61 and 67
The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9
The possibilities are:
05, 23, 41, 67, 89
05, 23, 47, 61, 89
05, 29, 41, 67, 83
05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9
The possibilities are:
02, 53, 41, 67, 89
02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7
The possibilities are:
02, 59, 41, 67, 83
02, 59, 47, 61, 83

There are a total of 8 sets of five primes

CrystalMath
CrystalMath
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December 2nd, 2020 at 4:51:41 PM permalink
Quote: unJon

Quote: CrystalMath

Quote: Gialmere


How many different sets of five primes can you form this way?



Your first assumption is incorrect:

You can make a card with 02 which would then let you make a 53 card or a 59 card.




d'oh
I heart Crystal Math.
Gialmere
Gialmere
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December 2nd, 2020 at 5:50:23 PM permalink
Quote: ThatDonGuy



0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).
The only primes < 100 that contain a 5 are 5, 53, and 59
4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2
If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29
The two-digit primes that begin with 4 are 41, 43, and 47
The two-digit primes that begin with 6 are 61 and 67
The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9
The possibilities are:
05, 23, 41, 67, 89
05, 23, 47, 61, 89
05, 29, 41, 67, 83
05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9
The possibilities are:
02, 53, 41, 67, 89
02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7
The possibilities are:
02, 59, 41, 67, 83
02, 59, 47, 61, 83

There are a total of 8 sets of five primes


Correct!

(Although, I spy a composite number typo in there.)
--------------------------------------

Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
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December 3rd, 2020 at 8:05:26 AM permalink


A fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?


That is, each roll is greater than or equal to the previous roll.
Have you tried 22 tonight? I said 22.
ThatDonGuy
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Gialmere
December 3rd, 2020 at 8:41:48 AM permalink
Quote: Gialmere

What is the probability that the sequence of rolls is non-decreasing?


This looks like a Markov chain problem - suitable more for computing than math.

Let P(a,b) be the probability of success after a rolls where the most recent roll is b
P(10,n) = 1 for all n
P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6
The solution is P(0,1)

Rolls123456
915 / 62 / 31 / 21 / 31 / 6
87 / 125 / 125 / 181 / 61 / 121 / 36
77 / 2735 / 2165 / 545 / 1081 / 541 / 216
67 / 7235 / 64835 / 1,2965 / 4325 / 1,2961 / 1,296
57 / 2167 / 4327 / 9727 / 2,5921 / 1,2961 / 7,776
477 / 7,77635 / 7,7767 / 3,8887 / 11,6647 / 46,6561 / 46,656
311 / 3,88855 / 46,6565 / 11,6641 / 7,7761 / 34,9921 / 279,936
2143 / 186,62455 / 186,62455 / 559,8725 / 186,6241 / 186,6241 / 1,679,616
11,001 / 5,038,848715 / 10,077,69655 / 2,519,42455 / 10,077,6965 / 5,038,8481 / 10,077,696

Result = 1,001 / 20,155,392 = about 1 / 20,135

Gialmere
Gialmere
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December 3rd, 2020 at 4:57:37 PM permalink
Quote: ThatDonGuy


This looks like a Markov chain problem - suitable more for computing than math.


Let P(a,b) be the probability of success after a rolls where the most recent roll is b
P(10,n) = 1 for all n
P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6
The solution is P(0,1)

Rolls123456
915 / 62 / 31 / 21 / 31 / 6
87 / 125 / 125 / 181 / 61 / 121 / 36
77 / 2735 / 2165 / 545 / 1081 / 541 / 216
67 / 7235 / 64835 / 1,2965 / 4325 / 1,2961 / 1,296
57 / 2167 / 4327 / 9727 / 2,5921 / 1,2961 / 7,776
477 / 7,77635 / 7,7767 / 3,8887 / 11,6647 / 46,6561 / 46,656
311 / 3,88855 / 46,6565 / 11,6641 / 7,7761 / 34,9921 / 279,936
2143 / 186,62455 / 186,62455 / 559,8725 / 186,6241 / 186,6241 / 1,679,616
11,001 / 5,038,848715 / 10,077,69655 / 2,519,42455 / 10,077,6965 / 5,038,8481 / 10,077,696

Result = 1,001 / 20,155,392 = about 1 / 20,135



Correct!

Very good.
-------------------------------

Have you tried 22 tonight? I said 22.
Ace2
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Gialmere
December 3rd, 2020 at 6:47:24 PM permalink
Quote: Gialmere

A fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?


That is, each roll is greater than or equal to the previous roll.

There's another way to calculate this. If we only count permutations that have at least one of each side 1-6, then the answer would be combin(9,5) = 126 (out of 6^10). This is because there are 126 ways to insert 5 dividers between 10 values so that 6 bins are created. This, for example, would represent 1123334556: xx/x/xxx/x/xx/x

However, a bin can contain zero values, so we add 1 for each of the 6 bins and "overcount" each side by 1. But not exactly since the value we are overcounting is a zero. x/xxx/x/xxxxxx/xxx/xx represents 2244444556, for instance.

Therefore the answer becomes combin(15,5) = 3,003 (out of 6^10). This is because there are 3,003 ways to insert 5 dividers between 16 values so that 6 bins are created.
Itís all about making that GTA
Gialmere
Gialmere
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December 4th, 2020 at 8:00:05 AM permalink


Oh no! You're driving your car through town when you see you're going to be stuck behind a big Budget Rental Truck at the next traffic light.

A Budget Rental Truck is 12 feet 8 inches tall (from the road to the top of the Van). The cab of the truck is 8 feet long and the van is 24 feet long (see diagram below).

If the front of the truck is 8 feet from the stoplight suspended 18 feet above the road, how far behind the truck must your car be so that you can still see the stoplight?

Assume your eyes are 4 feet above the road.


Have you tried 22 tonight? I said 22.
ThatDonGuy
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Gialmere
December 4th, 2020 at 8:26:19 AM permalink

The top of the truck is 5 1/3 feet below the stoplight, and 40 feet to the right of the stoplight
Let X be the distance you are behind the truck
Your eyes are 14 feet below the stoplight, and (40 + X) feet to the right of the stoplight
The minimum distance where you can see the stoplight is where the two height/length ratios are equal
40 / (5 1/3) = (40 + X) / 14
15 / 2 = (40 + X) / 14
210 = 80 + 2X
X = 65 feet


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