If a fair die is rolling 3 times, what room the odds of getting an also number on each of the an initial 2 rolls, and an strange number top top the 3rd roll?

I think the permutations formula is needed i.e. $n!/(n-r)!$ since order matters however I"m not sure if n is 3 or 6 and what would certainly r be?

Any assist would be much appreciated!

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Let"s calculate the probability, then convert that come odds.

On a same die, half the number are even and fifty percent the numbers are odd. So, the probability for a solitary roll of acquiring an also number or one odd number is $dfrac12$.

The probability because that a particular roll room unaffected by previous rolls, so us can use the product principle and also multiply probabilities because that each roll. Every roll has actually probability of $dfrac 1 2$ that obtaining the preferred result. So, we have:

$$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$

Now, the probability that that no happening is $$1-dfrac 1 8 = dfrac 7 8$$

So, the odds space 7:1 against the wanted outcome.

answer Sep 21 "18 in ~ 13:23

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I think friend might be over complicating things.

It has to be even on the first two, the probability the this is $frac12 imes frac12 = frac14.$

The probability of odd on the 3rd roll is additionally $frac12$ so your last probability is $frac18.$

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answered Sep 21 "18 in ~ 13:22

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Given a fair die, the probability of any identified sequence of 3 evens or odds is the same, namely, $1/2 imes 1/2 imes 1/2 = 1/8$. Therefore the odds are $7$ to $1$ against.

answer Sep 21 "18 in ~ 18:44

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Assuming the die is just marked even and also odd fairly than through numbers, there room eight orderings. They variety from every odd to every even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In her formula:


You are missing that friend don"t care around the assignment of the duplicates. So you need


The $n$ is the total variety of rolls, the $r$ is the number of either evens or odds (it"s symmetric). But to gain the total number of orderings, you require to include these:


Now substitute 3 for $n$.


Unrolling that, we get

$$frac3!3!0! + frac3!2!1! + frac3!1!2! + frac3!0!3!$$


$$1 + 3 + 3 + 1$$

So we have one every odds, three with one even, three with two evens, and one all even. That"s eight total.

Another way of thinking of this is that there is only one bespeak of every odd numbers or all also numbers while there are three places where the lone odd or also number deserve to be.

Of those eight, how countless fit her parameters? precisely one, OOE. So one in eight or $frac18$.

As rather have already noted, you can get the much much more easily by merely figuring the you have a one in two chance of acquiring the result you need for each roll. There"s three rolls, for this reason $(frac12)^3 = frac18$.

If you desire to act 1, 3, and 5 as various values and also 2, 4, and 6 as different values, you can. Yet it is much simpler to think of castle as just odd or even. Since you don"t want to try write this the end for $6^3 = 216$ orderings. And in the end, friend will acquire the same basic result. Girlfriend will have actually twenty-seven OOE orderings, i beg your pardon is again one eighth that the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $frac27216 = frac18$.

Permutations leads you under a more difficult path. It"s simpler to think simply in terms of probability or also ordering.