You roll a blue die and a yellow die. What is P(the sum of the dice is at least 6 | the blue die shows a 2)? Write fractions using a slash and reduce them to lowest terms.
@ybarrap
b/y 1/2
how did u get that
there is one blue and one yellow...that is 2. if you roll both its a 50 50 chance so 1/2
ok thnk u
yeahhhhh:)
what abt this?:
You roll two dice. Let event A be "The first die shows a 1 or a 6" and event B be "The second die shows an even number." What is P(A | B)?
In the first problem, if the blue die lands 2 then in order for the sum of the blue and yellow to be 6 or greater, the yellow needs to be a 4, 5 or 6, since 2 + 4, 2 +5 and 2 + 6 are all sums greater than or equal to 6, The chance that yellow is 4,5 or 6 is 3/6 or 1/2.
its not 1/3??
its 1/2 sleek-feathered one
No @MROAKS1432 the answer is 1/3
how did u get that
Because if P(A) = P(1 or 6) then: P(1) + P(6) and then you can get your answer: (1/6)+(1 / 6) = 2/6 =1/3
so it is 1/3. @ybarrap i don't really understand wat u said though
The correct answer is 1/3. Can I get a medal :)
it says that i already gave out a medal.. sorry
:( its ok but glad to help :)
thanks. can any of u guys open the attachment??? :(
My reply above was for your first problem. For the second problem P(A|B) = P(A and B)/P(B), but each are independent, so P(A|B) = P(A)P(B)/P(B) = P(A) = P(1) + P(6) = 1/6 + 1/6 = 1/3 as @AwesomeB stated
yay :)
ok what abt the ATTACHMENT above
ill repost it as a new question & then i'll tag u guys??
The attachment you said was a lot of random numbers and letters
I think that the ..rtf part of your doc name is causing problems, my browser is interpreting as a plain text, so the rtf-encoding is showing not the decoded rtf
ugh! so there's no way u guys can view it?? :(
no unless you can change it
change the name, remove one of the periods
ok.
even though I change the name it stays a an "rtf"..
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