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you use the fact that P(A) = 1- P(not A)
can u show me how to do each part?
let's do a
P(at least two 5s) = 1- P(not at least two 5s)
= 1- P(at most one 5)
so let's find P(at most one 5)
can you find this ?
I got 146/1296 for A
is that right???
I am not finding the answer, just trying to help
so P(at most one 5) = P(no 5s) + P(exactly one 5)
P(no 5) = \[(5/6)^4\]
P(exactly one 5) follows the binomial distribution, so \[4C1*(1/5)^1*(5/6)^4\]
when you subtract the sum of those from 1, you get the answer for a),
oops, the 1/5 should be 1/6
don't u need 0 5s as well?
I already did that, it is (5/6)^4
wait question is at least 25s
let me explain you that par again
P( at least two 5s) = P(exactly two) + P(exactly three)+P(exactly four)
oh yea I did that
it is a pain to calculate all those, so we use the rule P(event) = 1-P(complement of that event)
4C2(1/6)^2(5/6)^3 + 4C3(1/6)^3(5/6)^1 + 4C4(1/6)^4(5/6)^0
is that right?
what I did was 1- (5/6)^6 -4C1(1/6)(5/6)^3
for B, the complement is "no evens at all"
so 1-P(no evens) will be the answer
isn't it \[4C3(1/2)^3(1/2)^1 + 4C2(1/2)^2(1/2)^2 + 4C1 (1/2)^1(1/2)^3\]
you need to include 4C0(1/2)^4
everything else is correct?
az, just to let you know you probably know how to get the answers for all of this. But you also need to know that you are missing the point of what these problems are asking.
next one idk
so it can be 0 or 1 odd number?
4C1(1/2)^1 (1/2)^3 + 4C0(1/2)^0 (1/2)^4 ??
is that right
az, I need to go soon so let me leave you an advice
your method is logically straight forward, and as long as you don't mind calculating a lot of numbers you will get the answer so I have no doubt that you will get any of these wrong from what I have seen so far
can we go over the last two really quickly
but all of the problems here can be found a lot more easily using the fact P(A) = 1-P(notA)
4C1 (1/2)^1(1/2)^3 + 4C0 (1/2)^0 (1/2)^4 ????
so d is 1 - P (all of them being prime) and e is P(0) + P(1)
is that right? what I just put
what is it then
sorry, ask someone else since I need to go. :)
what is it then