if i throw two different dice at the same time what is the probability throwing a three?answer in fractions?

Question

put the answer in the lowest fraction terms

Answer 1

Upon rolling two dice, there are 36 possible outcomes:-
11--12--13--14--15--16
21--22--23--24--25--26
31--32--33--34--35--36
41--42--43--44--45--46
51--52--53--54--55--56
61--62--63--64--65--66

P(3) = 11/36

Answer 2

Well - you have a 1/6 chance of throwing a three with the one dice, and a 1/6 chance of through a three with the second dice. If you're only wanting the probability of getting ONE three, you'll add those those fractions - which gives you 2/6 which simplifies to 1/3.

Now - if you wanted the probability that BOTH die would be threes, you'd multiply the two - giving you 1/36.

:)

Oh - I just saw the person above me - hmm.. maybe I interpreted the question too literally. If it can be any combination of three, then his answer is correct. If you're talking about specifically rolling a three on one of the die (or both dice being three), then that's how my answer would apply.

Answer 3

maximum folk have stated 11/36 this is probable the magnificent answer. in spite of the fact that, as somebody has already stated, you may desire to be slightly greater particular with the wording of the question. in case you mean solely throwing a 'one' and in common terms one 'one' then the opportunities are 10/36 (or 5/18). In different words the potential of throwing 2 'ones' has been excluded. from time to time, an common way of doing those issues is to paintings out the completed opportunities (6 x 6) and subtract the probabilities or not throwing a 'one'. when you consider that there are 5 opportunities of failing to throw a 'one' on each and each cube, this provide us 5 x 5 (25) possibilites of failing to throw a 'one'. this means via deduction that there are 25/36 opportunities of not throwing a 'one' which provides us a 11/36 of throwing a minimum of one 'one'.

Answer 4

assuming fair dice

to get a sum of three you have 2 out of 36 possible outcomes for a probability of 1/18

to get a three on one face of one die you have 1/6 probability

the probability of getting only one three on a face of the two dice is 1/6*5/6 + 1/6*5/6 = 10/36.

the probability of double threes is 1/6 * 1/6 = 1/36

the probability of at least one three on a face is 11/36

Answer 5

do u mean throwing a 3 on both dice or 2 on one die and 1 on the other?

Answer 6

1/6 x 1/6=1/36

Answer 7

1/18.

36 possibilities. 2 possibilities to get 3, 2+1, 1+2.

2/36 = 1/18.