Question

Suppose you role four dice. Find the probabilities that you get 0, 1, 2, 3, and 4 threes. Give the probability to the nearest .0001

I guess the question is asking for 5 different trials.
(the question is as its given)

Answer 1

So you are rolling 4 dice at once? and you want the probabilities that you get (???)...please explain so that I may help :-)

Answer 2

I also think that combinations/permutations are used

Answer 3

roling 4 dice = rolling 1 die four times

Answer 4

For some reason that makes a lot more sense now to me then it did before...but why does the question state "Suppose you role 4 dice" instead of Suppose you role a dice 4 times...

Answer 5

So here it goes with the assumption that you role a die 4 times...

Answer 6

it was a test question, evil test

Answer 7

Wait wait wait you want the Probability that you get zero 3's (anything other than 3),
one 3 (3), two 3's (3, 3), three 3's (3, 3, 3), and four 3's (3, 3, 3, 3)

Answer 8

So I have 1 die. If I roll it the first time the probability that I'll get zero 3's is 5/6, do you agree?

Answer 9

maybe its asking:
How many times can you get a three if you roll it 4 times?
(i hate word problems)

Answer 10

P(zero threes) = 5/6 x 5/6 x 5/6 x 5/6 = .4822
P(one three) = 1/6 x 5/6 x 5/6 x 5/6 = .0965
P(two threes) = 1/6 x 1/6 x 5/6 x 5/6 = .0193
P(three threes) = 1/6 x 1/6 x 1/6 x 5/6 = .0039
P(four threes) = 1/6 x 1/6 x 1/6 x 1/6 = .0008

Answer 11

@beans1992 i think thats it!

Answer 12

thnx a bunch, you too @mashe

Answer 13

You know I think it's asking for one event, and in that event what is the probability that you get 0, 1, 2, 3 and 4 3's...
Okay so if you roll the 4 dice at once, the probability that you get zero 3's is the same as the probability that all of the dice have anything other than 3:
P(dice 1 has anything other than 3)*P(dice 2 has anything other than 3)*P(dice 3 has anything other than 3)*P(dice 4 has anything other than 3)=5/6+ 5/6+ 5/6 + 5/6
Notice order does not matter you could name any dice, dice one, dice two, dice three, and dice 4. Also, notice that you multiplied the probabilities together because of the rule that if you have two independent events A and B you have
P(A and B)= P(A) * P(B)
Now try the rest...

Answer 14

thx, really helped explain it better