[/quote]
,
is equal to
of each imon 
.
become entangled if and only if the original imons 
are entangled, and each copy 
members from party 
,
members from party 
,
members from party 
,
members from party 
,
or more people involved).
that if they were to be considered the number of members from the four parties, it is possible that, after some time, all of the kidnapped people belong to the same party, under the same rules imposed by the wizard.
at 
meets sides 
and 
again at 
and 
respectively. Let 
be the circumcentre of 
is perpendicular to 
.
for which 
is the square of an integer
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
such that 
.
,
.
.
has 
,
,
.
.
Y by RainbowJessa, Sharkship, RollingPanda4616, huajun78, PikaPika999, HacheB2031, GodGodGodGodGoose
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! 
Please hide solutions and answers, hints are fine though!
Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!
Please hide solutions and answers, hints are fine though!
Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!
This post has been edited 7 times. Last edited by Leeoz, Apr 14, 2025, 4:51 AM
and 
.
So, 
.
are 6 cases, which add up t 
.
contribute a total of 
.
contribute 
.
contribute 
.
and 
are the same number, as are 
and 
,
and 
,
.
in the product (prime factorization) although it's still kind of a bash
Since each is relatively prime, I can do the simple calculation 
it must have the die rolls 
and 
and 
and 
This gives 
and 
This gives 
gives 
The other person made a mistake
therefore Thursday
so it will be a Thursday.
jackpot! However, a ticket costs 
.
chance of gaining 
chance of losing 1 dollar. By using the formula for expected value, you are expected to gain 
.
,
,
cents.
then we want the probability of rain 6 times and not rain 1 time so we just have 50% for each of them to get 
to get 
but now we just multiply 7 and 1/128 to get 
ways to choose the 
.
.
which equals 
.
equally likely possible outcomes, and only 
of the outcomes satisfy this condition. The probability is ![\[\frac{\binom71}{128}=\boxed{\frac7{128}}.\]](http://latex.artofproblemsolving.com/d/a/b/dab9317f5be3000cb2fbee576529282f80bf0774.png)
?
