Melody

Hi Red,

Welcome back  

I certainly do not mind if you throw a daily riddle our way.

I miss the fun that we used to have here.     

If you can reinvent some of it I think that would be nice :))

There are 210 students in a twelfth grade high school class. 90 of these students have at least one sister and 105 have at least one brother. Out of these, there are 45 who have at least one sister and one brother.

Let A be the event that a randomly selected student in the class has a sister and B be the event that the student has a brother. Based on this information, answer the following questions.

What is P(A), the probability that the student has a sister? _____    \(\frac{90}{210}=\frac{3}{7}\)

What is P(B), the probability that the student has a brother?_____   \(\frac{105}{210}=\frac{1}{2}\)

What is P(A and B), the probability that the student has a sister and a brother? _____  \(\frac{45}{210}=\frac{3}{14}\)

What is P(B | A), the conditional probability that the student has a brother given that he or she has a sister?   \(\frac{45}{90}=\frac{1}{2}\)

Is P(B | A)=P(B)?   YES   Are the events A and B independent?   Yes I think they are independent.

Here is the corresponding Venn diagram.

The town of Hamlet has  3 people for each horse, 4 sheep for each cow, and  3 ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

Answer Choices:

A. 41

B. 47

C. 59

D. 61

E. 66

1Horse, 3People, 9ducks = some multiple of 13

1Cow, 4 sheep = some multiple of 5

So just like Rom said the answer will be such some multiple of     13H+5C

Now the multiples of 13 are   0, 13, 26, 39, 42, 55, 68 and 68 is bigger than any choice so I do not need to go further

5C will end in 0 or 5 so I can add  0 or 5 to the last digit of the multiples of 13 above

41=26+15

47= ???  to get an answer of 47 the multiple of 13 would have to end in 7 or 2 and none do so it is not a possible answer.

I got the same answer,

Here is my take.

A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Let the length of the line be 1 unit.

Let the lengths of the segments be  x, 1-y, and y-x   units.      (if you add these they equal 1)

Now all these segments must be between 0 and 1 unit

So I will graph this and the area will give the full population of possible cuts.

Represented on this contour map in dark green

Now lets see how much of this is included if the peices can form a triangle.

To make a triangle the sum of the smaller peices has to be bigger than the length of the longest one.

so all three of the following conditions must be met.

\(x+1-y>y-x, \qquad x+y-x>1-y, \qquad 1-y+y-x>x\\ \text{These simplify to }\\ y 0.5, \qquad x<0.5\)   

So now I can plot these over the population and see what happens.

I can see that a quarter of the population triangle is included so

The probablility that the peices will be able to form a triangle is 1/4

(I keep editing this because some of what I write keeps getting deleted. There must be more elements here than web2.0 can happily cope with. I will see what happens this time but if it is not complete you will have to join the dots for youselves :/)

Don't know, I've answered this question before, more then once, but I never find it  simple....

mmmmm  I could just chase down an old answer .... would that be cheating?  That is rhetorical,    LOL.

Why do you want another set of eyes Rom?  

The most important thing to implant into your brain is that the first derivative at a point IS the gradient of the tangent at that point.  They are the same thing!

Take a loook at this​

Yes PM is right. More info is needed.

Thanks Alan,

I had a super stupid error in mine but I have corrected it and I think our answers are the same now 

Here is the graph of  Toatal area in terms of x

Remember :

6x metres is cut off for the TRIANGLE

and   

5-6x metres is used for the square

(edited to fit with the corrected answer above)