#repost cuz need explanations
77 messages · Page 1 of 1 (latest)
I'm pretty sure for the first one, you can just multiple the proabbilities
so find the probability of both 20 c coins landing heads
which you can also find by multiplication
the probability of landing heads assuming it's fair coin is 50%
so it's 1/2
since you do this twice, that's 1/2 * 1/2
which is 1/4
now for landing at least one 10 c coins
you could either create a sample space or use the binomial distrbution equation or whatever it's called
something like this
and checking which ones have at least one head
which is 7/8
and now thinking about it
you don't really need a sample space for this
the only case where "at least one of the heads landing" tails is not true
is when when land all tails
which only happens once in a sample space like this
and in general the number of outcomes in this case will be 2^n (since each time you flip a coin, two more possibilities can happen for each possibility)
so the first flip is only 2 outcomes
the second flip, there is 2 more outcomes, that will branch off of those two initial outcomes
so 2 will branch off of the first outcome
2 will branch off of the second outcome
hence we now have 4
and so on
so it's just multiplying by 2 each flip
so the general equation could be written as ```
1/(2^n - 1)
anyways the probability is 7/8 for at least landing a head on one of the three 10 c flips)
the probability of both happening is just the product of the two probabilities
this is assuming the two events are independent
which you can probably assume in this case since there isn't much reason to believe the events aren't dependent considering it's coin flips
so 1/4 * 7/8
7 / 32
for the second one you can kind of imagine a right triangle
and the reason we want to set this up
is because we want to find that base length
because the lid is slightly bigger than the bottom of the bucket
you can think of the lid as being equal to the base
mhm
plus some value on the left and right side
which are presumably equal
so let's say that the "red" line is equal to "x"
using the pythagorean theorem we can set up
$$x^2 + 30^2 = 31^2$$
MoistBrain
MoistBrain
and we can think of the lid as 18 + x + x
since it's a bit bigger than the base by "x" amount
on the left and right side
which is why we add it twice
so that'll give you
$$18 + 2\sqrt{31^2 - 30^2}$$
MoistBrain
MoistBrain
3rd one?
you add the vectors up
you can do that by putting the tail of one vector on the head of the other
and seeing where it ends up
so you can like move the tail of vector "c" to the head of vector "p"
just kind of do it more accurately than this lol