dice probability helpp!!?
if you have a 6 side dice and you roll the dice 3 times what are the probability of getting a 3 ? i need to know you get the answer not the answers it seft please help
- MerlynLv 71 decade agoFavorite Answer
Let X be the number of "3's" observed in three rolls of the die.
X has the binomial distribution with n= 3 trials and success probability of 1/6.
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(n!(n-1)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
plug and chug...
P(X = 0) = no threes = 0.5787
P(X = 1) = 1 three in the three rolls = 0.3472
P(X = 2) = 2 threes in three rolls = 0.0694
P(X = 3) = all three rolls are a three = 0.00462
- 1 decade ago
Ok...i m assuming u only want a 3 or 3 for all three throws?
For only one 3.
Each throw, the probability of getting a 3 is 1/6
So if you want STRICTLY a 3 then it would be
(1/6) (times) (5/6) (times) (5/6) (times) (3!/2!)
The first one sixth is to get a 3 and the other 2 5/6 is to get anthing else BUT 3. and lastly, the 3!/2! is to use permutations to find the different combinations (i.e. at which try u get a 3)
BUT...if you question is 3 on all three tries then its (1/6) to the power of three...cos u need 3 at all three tries.
General rule of thumb: and = times
or = plus
SO in this case since the "story" goes getting a 3 AND a 3 AND another 3...so you have to take the probablility of getting a three and mulitply it with itself 3 times...
Also, another way of looking at your question...is in the three tries then you get 3 as a total? Like each time you throw its a 1...and so and 1 for 3 times...gives u a sum of 3...
If the question is like that then u need to take 1/6 and put it to the power of 3...
If the 3 u are talking about involves all the possibilities i talked about above?? Then just add the probability form the 3 cases together...yup
All the bestSource(s): I studied this in school...yeah so my grade literally depends on it...
- mangerLv 44 years ago
a) to get a sum of 6, you may have a million,5 or 5,a million or 4,2 or 2,4 or 3,3 = 5 achievable variations. There are a entire of 6*6 = 36 variations, so your odds are 5/36 b)To get a made from 12, you may have 2,6 or 6,2 or 3,4 or 4,3 =>4/36 = a million/9 c) a million/6 d)a million/6 you won't be in a position to roll a 5 and get a made from 12, nor are you able to get the two a sum of 6 and a made from 12. (together unique)