What is the probability that, on three rolls

[prengasn]

What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?

This problem is taken from manhattan GMAT/Word translations and is solved as follows:

The probability that on all 3 rolls the die will not yield a 6 is 5/6*5/6*5/6=125/216
P(at least one will be 6)= 1-125/216.

Why can't we solve it as folows: P(we get one 6)=1/6, thus P( we get at least one 6 on three rolls)=1/6*1/6*1/6.

We can multiply Ps of the event not happening, but we can't multipliy Ps of the events happening???

I appreciate an explanation

[Kweku.Amoako]

The probability of at least a 6 on 3 rolls is the inverse of the probability of no 6 at all.

to solve for P(at least one 6) directly you will have to consider these events. Let Y represent a getting a six and N represent not getting a 6. Then after 3 rolls you get and of these results
N N N ( NO 6 )
N N Y ( one 6)
N Y N ( one 6)
N Y Y (two 6s)
Y N N (one 6)
Y N Y (two 6s)
Y Y N (two 6s)
Y Y Y (three 6s)

If you multiply (1/6) * (1*6) * (1/6) that would be the probabilty that you get only one 6. You will be ignoring the possibility of two 6s and three 6s. As you can see above there are far much more scenarios to consider for at least one 6 than for no 6.

[prengasn]

Thank you again

[RonPurewal]

We can multiply Ps of the event not happening, but we can't multipliy Ps of the events happening???

two things.

(1) you MULTIPLY if you want ALL of the given events to occur together.
so, if you wanted 6's on ALL THREE rolls of the die, then (1/6) x (1/6) x (1/6) is exactly how you would find that particular probability.

(2) you should CHECK PROBABILITIES TO SEE WHETHER THEY MAKE SENSE.
probability can be - and often is - hard. however, probability is also extremely intuitive; all of us have some sort of notion about percentage likelihood.
in that light:
1/6 x 1/6 x 1/6 is WAY smaller than 1/6.
the chance of success IN THE FIRST TRY, ALONE, is already 1/6.
therefore, any probably LESS than 1/6, for the probability of success in MULTIPLE tries, makes no sense at all.
this would be akin to having a smaller chance of success in the lottery the more tickets you buy.

always think about whether probabilities make sense!

[vananh.nguyen.vn]

The probability of at least a 6 on 3 rolls is the inverse of the probability of no 6 at all.

If you multiply (1/6) * (1*6) * (1/6) that would be the probabilty that you get only one 6. You will be ignoring the possibility of two 6s and three 6s. As you can see above there are far much more scenarios to consider for at least one 6 than for no 6.

I disagree w/t your explanation

If you multiply (1/6)*(1/6)*(1/6) --> the prop. of getting three 6, not one.

[vananh.nguyen.vn]

Here is my view. Tell me why i am wrong:

+ The Prop of getting one six: (1/6)*(5/6)*(5/6)=25/216

+ The Prop of getting two six: (1/6)*(1/6)*(5/6)=5/216

+ The Prop of getting three six: (1/6)*(1/6)*(1/6)=1/216

So the final result would be the sum of the three = 31/216, not 91/216 like the book

Thank you

[RonPurewal]

Here is my view. Tell me why i am wrong:

+ The Prop of getting one six: (1/6)*(5/6)*(5/6)=25/216

+ The Prop of getting two six: (1/6)*(1/6)*(5/6)=5/216

+ The Prop of getting three six: (1/6)*(1/6)*(1/6)=1/216

So the final result would be the sum of the three = 31/216, not 91/216 like the book

Thank you

this analysis is incorrect because you haven't considered all the cases. when you multiply together probabilities, the ORDER of the events is ALWAYS included in the consideration, whether you like it or not.

so, your listed probability of getting one six is actually the specific probability of getting a six, THEN not getting a six, THEN again not getting a six.

likewise, your listed probability of getting 2 sixes is actually the specific probability of getting a six, THEN getting another six, THEN not getting a six.

if you're going to go this route, you have to account for the other probabilities.
so in your analysis for "one six" you have to add together the following three probabilities:
1/6 x 5/6 x 5/6
5/6 x 1/6 x 5/6
5/6 x 5/6 x 1/6
... and in your analysis for "two sixes" you have to add together the following three probabilities:
1/6 x 1/6 x 5/6
1/6 x 5/6 x 1/6
5/6 x 1/6 x 1/6
if you have these together to the probability that you have (correctly) calculated for three sixes, then you will get the correct aggregate probability.

in any case, the opposite event here (not getting any sixes) has a much easier probability to calculate, so it's easier to do the problem that way. however, the best approach to any problem is to accumulate as many solutions as possible, so at least you are doing that.