I want to revitalize this thread, so I will start with a variation of my first riddle. Here it is:
FakeName6 wrote:
Here is a probability problem of my own invention:
A man has a drawer with 150 socks: 50 red, 40 orange, 30 yellow, 20 green, and 10 blue. He takes six socks out. What is the chance that at least two of them are the same color?
My new one, unlike this earlier one (the answer was 100%, for those who missed it), is not a trick question. It involves real statistics. I am not totally sure that I got it right, so you guys might get me on it. Here it is:
The situation is the same as the first one. Except now you just pull two socks.
Simple, right? Or is it....
The chance of drawing a red sock followed by another red is (50/150) x (49/149). Add to that the chance of an orange followed by another orange (40/150) x (39/149). Add to that the chances of other-coloured pairs worked out similarly. The numerical answer is left as an exercise for the student.
.2389739 ect
"etc" is ambiguous. Wrong answer.
I want to revitalize this thread, so I will start with a variation of my first riddle. Here it is:
FakeName6 wrote:
Here is a probability problem of my own invention:
A man has a drawer with 150 socks: 50 red, 40 orange, 30 yellow, 20 green, and 10 blue. He takes six socks out. What is the chance that at least two of them are the same color?
My new one, unlike this earlier one (the answer was 100%, for those who missed it), is not a trick question. It involves real statistics. I am not totally sure that I got it right, so you guys might get me on it. Here it is:
The situation is the same as the first one. Except now you just pull two socks.
Simple, right? Or is it....
The chance of drawing a red sock followed by another red is (50/150) x (49/149). Add to that the chance of an orange followed by another orange (40/150) x (39/149). Add to that the chances of other-coloured pairs worked out similarly. The numerical answer is left as an exercise for the student.
.2389739 ect