## Poll

 I love math! 20 votes (46.51%) Math is great. 14 votes (32.55%) My religion is mathology. 6 votes (13.95%) Women didn't speak to me until I was 30. 2 votes (4.65%) Total eclipse reminder -- 04/08/2024 12 votes (27.9%) I steal cutlery from restaurants. 3 votes (6.97%) I should just say what's on my mind. 6 votes (13.95%) Who makes up these awful names for pandas? 5 votes (11.62%) I like to touch my face. 12 votes (27.9%) Pork chops and apple sauce. 9 votes (20.93%)

43 members have voted

camapl
Joined: Jun 22, 2010
• Posts: 420
July 19th, 2022 at 4:08:40 AM permalink
Quote: ThatDonGuy

Since nobody else has published one today, here's a (relatively) easy one for Monday:

A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?

Thanks for Monday’s question, although I’m answering on Tuesday!

Considering your example, where N = 4, as you’ve already done the math, boys have a sister 1/4 + 3/8 + 1/4 = 7/8 or 14/16 of the time, and girls have a sister 3/8 + 1/4 + 1/16 = 11/16 of the time. If you prefer a simpler example, consider N = 2. You only have 2 girls (girl with a sister) 1/2 * 1/2 = 1/4 of the time, while a boy has a sister (1/2 * 1/2) + (1/2 * 1/2) = 1/2 of the time. Answer: more boys than girls have a sister. Consequently, more girls than boys have a brother.
* Actual results may vary.
Gialmere
Joined: Nov 26, 2018
• Posts: 2589
July 19th, 2022 at 8:57:38 AM permalink
It's toughie Tuesday. Back to goats...

The cross section of a shed is a 10 foot by 10 foot square. The shed is located in the center of an open, level field.

A goat is tethered to one corner of the shed by a forty foot rope. The goat cannot enter the shed.

What is the area over which the goat can graze?

Picture not to scale.
Have you tried 22 tonight? I said 22.
ChesterDog
Joined: Jul 26, 2010
• Posts: 1136
Thanks for this post from:
July 19th, 2022 at 4:38:59 PM permalink
Quote: Gialmere

It's toughie Tuesday. Back to goats...

The cross section of a shed is a 10 foot by 10 foot square. The shed is located in the center of an open, level field.

A goat is tethered to one corner of the shed by a forty foot rope. The goat cannot enter the shed.

What is the area over which the goat can graze?

Picture not to scale.

I get about 4,847.07 square feet.

This is the result of the following expression:

Wizard
Joined: Oct 14, 2009
• Posts: 25287
July 19th, 2022 at 4:40:17 PM permalink
Quote: Gialmere

What is the area over which the goat can graze?

I'm getting the answer below, but my confidence is not very high. If I'm right, I'll provide a solution.

4891.25444 square feet

Hoping to get a reply to my reply to my reply on the probability of a sister problem.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5680
July 19th, 2022 at 4:57:06 PM permalink
Quote: camapl

Quote: ThatDonGuy

Since nobody else has published one today, here's a (relatively) easy one for Monday:

A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?

Thanks for Monday’s question, although I’m answering on Tuesday!

Considering your example, where N = 4, as you’ve already done the math, boys have a sister 1/4 + 3/8 + 1/4 = 7/8 or 14/16 of the time, and girls have a sister 3/8 + 1/4 + 1/16 = 11/16 of the time. If you prefer a simpler example, consider N = 2. You only have 2 girls (girl with a sister) 1/2 * 1/2 = 1/4 of the time, while a boy has a sister (1/2 * 1/2) + (1/2 * 1/2) = 1/2 of the time. Answer: more boys than girls have a sister. Consequently, more girls than boys have a brother.

You are counting how many families with boys versus how families with girls have sisters. I am looking for the actual numbers of boys and girls with sisters.

For example, your 1/16 value for the girls forgets that this is four girls that have one or more sisters.

Wizard
Joined: Oct 14, 2009
• Posts: 25287
Thanks for this post from:
July 19th, 2022 at 5:19:48 PM permalink
Quote: ChesterDog

I get about 4,847.07 square feet.

This is the result of the following expression:

I see we're a little off. For now, let me throw out this diagram for purposes of discussion.

Here is some breakdown of my answer.

Area Size qty Total
Big quarter slices 1256.637061 3 3769.911184
Small slices 457.5939873 2 915.187975
Triangle 103.0776406 2 206.155281
Total 4891.254440
Last edited by: Wizard on Jul 19, 2022
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ChesterDog
Joined: Jul 26, 2010
• Posts: 1136
July 19th, 2022 at 6:34:10 PM permalink
Quote: Wizard

Quote: ChesterDog

I get about 4,847.07 square feet.

This is the result of the following expression:

I see we're a little off. For now, let me throw out this diagram for purposes of discussion.

Here is some breakdown of my answer.

Area Size qty Total
Big quarter slices 1256.637061 3 3769.911184
Small slices 457.5939873 2 915.187975
Triangle 103.0776406 2 206.155281
Total 4891.254440

That's a good diagram!

I found the area of the upper yellow triangle by doing half the product of its base and height. I used 10 feet as the base.

To calculate its height, I used point A as the origin and found the intersection of the line y = x - 10 with the circle x2 + y2 = 900.

For the intersection point at point D I got x = -15.615528 and y = -25.615528. I used 15.615528 as the yellow triangle's height. So, I found one yellow triangle's area is (1/2)(10)(15.615528) = 78.077641
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5680
Thanks for this post from:
July 19th, 2022 at 8:22:45 PM permalink
Quote: ChesterDog

That's a good diagram!

I found the area of the upper yellow triangle by doing half the product of its base and height. I used 10 feet as the base.

To calculate its height, I used point A as the origin and found the intersection of the line y = x - 10 with the circle x2 + y2 = 900.

For the intersection point at point D I got x = -15.615528 and y = -25.615528. I used 15.615528 as the yellow triangle's height. So, I found one yellow triangle's area is (1/2)(10)(15.615528) = 78.077641

Here's what I get:
AC = 10, AD = 30, and angle ACD = 135 degrees
Law of Sines: (sin ADC) / 10 = (sin ACD) / 30, so sin ADC = (sin ACD) / 3 = sqrt(2) / 6; also, cos^2 ACD = 1 - sin^2 ACD = 17/18, so cos ACD = sqrt(34) / 6
Angle CAD = 180 - (ACD + 135) = 45 - ACD, so sin CAD = sin 45 cos ACD - cos 45 sin ACD = (sqrt(2) / 2) (cos ACD - sin ACD)
= (sqrt(2) / 2) (sqrt(34) - sqrt(2)) / 6
= (2 sqrt(17) - 2) / 12 = (sqrt(17) - 1) / 6
Area of CAD = 1/2 AC AD sin CAD = 150 (sqrt(17) - 1) / 6 = 25 (sqrt(17) - 1), which matches ChesterDog's number
camapl
Joined: Jun 22, 2010
• Posts: 420
July 20th, 2022 at 1:11:26 AM permalink
Quote: ThatDonGuy

Quote: camapl

Quote: ThatDonGuy

Since nobody else has published one today, here's a (relatively) easy one for Monday:

A recent census came up with some surprising numbers.
For all positive integers N, if you look at all of the families with N children, it turns out that the number of families with 0, 1, 2, ..., N boys is proportional to what it would be if the probability of any particular child being a boy was 1/2 (which, theoretically, it is, at least if you assume that all of the males are XY).
For example, among all of the families with 4 children, 1/16 of them have four boys, 1/4 have three boys and a girl, 3/8 have two boys and two girls, 1/4 have one boy and three girls, and 1/16 have four girls.
For any particular value of N, do more of the boys in N-child familes or more of the girls in those families have any sisters?

Thanks for Monday’s question, although I’m answering on Tuesday!

Considering your example, where N = 4, as you’ve already done the math, boys have a sister 1/4 + 3/8 + 1/4 = 7/8 or 14/16 of the time, and girls have a sister 3/8 + 1/4 + 1/16 = 11/16 of the time. If you prefer a simpler example, consider N = 2. You only have 2 girls (girl with a sister) 1/2 * 1/2 = 1/4 of the time, while a boy has a sister (1/2 * 1/2) + (1/2 * 1/2) = 1/2 of the time. Answer: more boys than girls have a sister. Consequently, more girls than boys have a brother.

You are counting how many families with boys versus how families with girls have sisters. I am looking for the actual numbers of boys and girls with sisters.

For example, your 1/16 value for the girls forgets that this is four girls that have one or more sisters.

For N = 4, boys have a sister 3 * 1/4 + 2 * 3/8 + 1 * 1/4 = 1-3/4 of the time, and girls have a sister 2 * 3/8 + 3 * 1/4 + 4 * 1/16 = 1-3/4 of the time. Thus, an equal number of boys and girls have a sister, as Mr. Wizard answered first.

For N=2, boys have a sister 1/2 the time (no change), while girls have a sister 2 * 1/2 * 1/2 = 1/2 the time as well.
* Actual results may vary.
Wizard
Joined: Oct 14, 2009
• Posts: 25287
Thanks for this post from:
July 20th, 2022 at 5:59:43 AM permalink
As to the goat problem, I now agree with CD. I had a couple of mistakes, including not subtracting out half the size of the shed.

Area Size qty Total
Big quarter slices 1256.637061 3 3769.911184
Small slices 460.5026797 2 921.005359
Triangle 103.0776406 2 206.155281
Less quarter square -25 2 -50.000000
Total 4847.071825

I also computed angle ADC incorrectly. It is atan(sqrt(2)/sqrt(34)).
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan