anonymous
  • anonymous
A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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yuki
  • yuki
you use the fact that P(A) = 1- P(not A)
anonymous
  • anonymous
can u show me how to do each part?
yuki
  • yuki
let's do a

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yuki
  • yuki
P(at least two 5s) = 1- P(not at least two 5s)
yuki
  • yuki
= 1- P(at most one 5)
yuki
  • yuki
so let's find P(at most one 5)
yuki
  • yuki
can you find this ?
anonymous
  • anonymous
um
anonymous
  • anonymous
I got 146/1296 for A
anonymous
  • anonymous
is that right???
yuki
  • yuki
I am not finding the answer, just trying to help
yuki
  • yuki
so P(at most one 5) = P(no 5s) + P(exactly one 5)
yuki
  • yuki
P(no 5) = \[(5/6)^4\]
yuki
  • yuki
P(exactly one 5) follows the binomial distribution, so \[4C1*(1/5)^1*(5/6)^4\]
yuki
  • yuki
when you subtract the sum of those from 1, you get the answer for a),
anonymous
  • anonymous
wait
yuki
  • yuki
oops, the 1/5 should be 1/6
anonymous
  • anonymous
don't u need 0 5s as well?
yuki
  • yuki
I already did that, it is (5/6)^4
anonymous
  • anonymous
wait question is at least 25s
yuki
  • yuki
let me explain you that par again
yuki
  • yuki
P( at least two 5s) = P(exactly two) + P(exactly three)+P(exactly four)
anonymous
  • anonymous
oh yea I did that
yuki
  • yuki
it is a pain to calculate all those, so we use the rule P(event) = 1-P(complement of that event)
anonymous
  • anonymous
4C2(1/6)^2(5/6)^3 + 4C3(1/6)^3(5/6)^1 + 4C4(1/6)^4(5/6)^0
anonymous
  • anonymous
is that right?
yuki
  • yuki
Yes.
anonymous
  • anonymous
k
anonymous
  • anonymous
B
yuki
  • yuki
what I did was 1- (5/6)^6 -4C1(1/6)(5/6)^3
yuki
  • yuki
for B, the complement is "no evens at all"
yuki
  • yuki
so 1-P(no evens) will be the answer
anonymous
  • anonymous
isn't it \[4C3(1/2)^3(1/2)^1 + 4C2(1/2)^2(1/2)^2 + 4C1 (1/2)^1(1/2)^3\]
anonymous
  • anonymous
???
yuki
  • yuki
you need to include 4C0(1/2)^4
anonymous
  • anonymous
oh yea
anonymous
  • anonymous
everything else is correct?
yuki
  • yuki
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.
yuki
  • yuki
Yes.
anonymous
  • anonymous
next one idk
anonymous
  • anonymous
so it can be 0 or 1 odd number?
anonymous
  • anonymous
??
yuki
  • yuki
yep
anonymous
  • anonymous
4C1(1/2)^1 (1/2)^3 + 4C0(1/2)^0 (1/2)^4 ??
anonymous
  • anonymous
is that right
yuki
  • yuki
yes
yuki
  • yuki
az, I need to go soon so let me leave you an advice
yuki
  • yuki
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
anonymous
  • anonymous
wait
anonymous
  • anonymous
can we go over the last two really quickly
yuki
  • yuki
but all of the problems here can be found a lot more easily using the fact P(A) = 1-P(notA)
anonymous
  • anonymous
4C1 (1/2)^1(1/2)^3 + 4C0 (1/2)^0 (1/2)^4 ????
yuki
  • yuki
so d is 1 - P (all of them being prime) and e is P(0) + P(1)
anonymous
  • anonymous
is that right? what I just put
yuki
  • yuki
nope
anonymous
  • anonymous
what is it then
yuki
  • yuki
sorry, ask someone else since I need to go. :)
anonymous
  • anonymous
????
anonymous
  • anonymous
what is it then

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