Melody

Good spotting guest.

I found my error, I did not read the question properly.    

P( at least 2 make it into the bin) 

= P(3)+P(4)+P(5)+P(6)

=1- [P(0)+P(1)]

Using the probabilities that I found before:

P(0)=\((\frac{2}{3})^6=\frac{64}{729}\)

\(P(1)=\frac{64}{263}=\frac{192}{729}\)

\(1-[P(1)+P(2)]=1-\frac{64+192}{729}\\ 1-[P(1)+P(2)]=1-\frac{256}{729}\\ 1-[P(1)+P(2)]=\frac{729-256}{729}\\ 1-[P(1)+P(2)]=\frac{473}{729}\\ \)

Now our answers agree.

Coding:

1-[P(1)+P(2)]=1-\frac{64+192}{729}\\
1-[P(1)+P(2)]=1-\frac{256}{729}\\
1-[P(1)+P(2)]=\frac{729-256}{729}\\
1-[P(1)+P(2)]=\frac{473}{729}\\
 

ok, another new one. :)   (make sure you fully understand the last one first though)

y=  0.5sin(2x) -1

1) midline   dashed line

2) topline (my term, not formal)           very light line    Hint:  it is a line, the equation needs any equal sign.

3) bottom line (my term, not formal)     very light line

4) wavelength (show a little working)

5) A quarter of the wavelength (this will help with choosing the scale on the x-axis)

NOTE: At this point you should have a rough sketch started

6) What would be sensible points for the x-axis? 

       Easiest to see this from your rough sketch that you are already making.

7) Will it start on the midline, above or under, why? 

                      So far you have just done this from memory of curve shape which is great. 

                       But, but you can just substitute.

                      when x=0    y=-2cos(0)+3 = ?

8) Will is start going down or going up, why?

9) y intercept

10) plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

                  (I mean maxima, minima, and points on the midline)

11) Graph it from x=0 to x= wavelength

12) what is the first maxima? (read from graph)

13) what is the first minima?

14) What points are on the midline?

Copy all these questions and answer on your own post. 

You can just copy my questions and answer after them if you want. (like last time)

You can check your sketch and answers with Desmos.  Make the x axis step equal to 1/4 of the wavelength.

Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability 211/243. If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin?

Edit:

NOTE: I have given the correct answer to the WRONG question.  There is another post below where I have answered the question that was actually asked    

If I shoot 5 bits the prob that they all miss is   

\( P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\ P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\ so\\ F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\ F=\frac{2}{3}\qquad S=\frac{1}{3} \)

\( P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\ P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\ P(none\; in)=(\frac{2}{3})^6\\ P(none\; in)=\frac{64}{729} \\~\\ P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\ P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\ P(one\; in)=2\;*\;(\frac{32}{243})\\ P(one\; in)=\frac{64}{243}\\ \\~\\ P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\ P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\ P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\ P(two\; in)= \frac{80}{243}\\ \\~\\ P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\ P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\ P(\text{no more than 2 in)}=\frac{496}{729}\\ \)

\(P(\text{no more than 2 in)}\approx 0.68\)

There could easily be careless mistakes, maybe even careless logic mistakes, but the underlying idea is sound.

My answer almost agrees with the second guest, and much of the logic is in agreement with Gino.

So we all had similar ideas.

Coding:
P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\
P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\
so\\
F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\
F=\frac{2}{3}\qquad S=\frac{1}{3}

P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\
P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\
P(none\; in)=(\frac{2}{3})^6\\
P(none\; in)=\frac{64}{729}
\\~\\
P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\
P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\
P(one\; in)=2\;*\;(\frac{32}{243})\\
P(one\; in)=\frac{64}{243}\\
\\~\\
P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\
P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\
P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\
P(two\; in)= \frac{80}{243}\\

\\~\\
P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\
P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\
P(\text{no more than 2 in)}=\frac{496}{729}\\

Not typos, the latex has just been omitted and guest is too lazy to even notice.

Thanks Heureka :)

It is really good to thank people but do not assume that all answers are correct.

Really!   I am astonished!     

I gave you a new answer under Gino's but it is probably not correct.

From a standard deck of cards, the four Aces, four 2's, four 3's, and four 4's are drawn, forming a smaller deck of 16 cards. All 16 cards are dealt at random to four players so that each player gets four cards. What is the probability that each player gets an Ace?

I have not looked at Geno's answer, I just decided to think about it independently.

I certainly will not promise mine is correct.

There are   16C4*12C4*8C4*4C4 ways to choose the hands but I this will be double counting so I think I need to divide by

4!

=63063000/4!

=2627625   total selection of hands (maybe)

Take all the Aces out and give three cards to each

(12C3*9C3*6C3) /4!*4!  maybe = 369600

369600/2627625

369600/2627625 = 0.1406593406593407 = 64/455     (probably wrong)

----------------------

Tell us when you know the answer, preferably with working:)

I looked at one, although my answer may not be correct

The other already seemed to be looked at. I did not check for accuracy.