A die is thrown three times. Show the probability of scoring a $2$ on just one occasion in the probability tree diagram and find the probability.

My Approach:

When a die is thrown once, the sample space is $6$. When it is thrown twice, the sample space is $36$ and when the die is thrown thrice, the sample space is $216$. But, how could I adjust the tree diagram with $216$ sample space in one piece of paper.

Moreover, I asked this question to my teacher at school, he, too could not draw the tree diagram. However, he got the answer as $\frac {25}{72}$, but the answer given in my book is $\frac {1}{8}$.

Please, help me to solve this.

  • 6
    $\begingroup$ "The sample space is 6." No. The size (cardinality) of the sample space is 6. $\endgroup$ – symplectomorphic Aug 21 '16 at 16:34

Why bother drawing everything from 1 to 6 if all you care about is if it's 2 or not. So you really have $2$ and $\overline{2}$

The three cases of rolling a single 2 in all 3 throws can be represented as

$\overline{2}$ - $\overline{2}$ - $2$

$\overline{2}$ - $2$ - $\overline{2}$

$2$ - $\overline{2}$ - $\overline{2}$

Is it ${1\over8}$ or ${25\over72}$?

To find the overall probability of all 3 cases, remember the 2 rules for such trees:

  1. going along a branch means multiplying probabilities and the result is the probability of the entire branch to happen
  2. adding the probabilities for several branches is done by adding them together

The single probabilities are $$P\big(2\big) = {1\over6} \qquad P\big(\overline{2}\big) = 1-P\big(2\big) = {5\over6}$$

The probability for the first branch is $$P\big(\overline{2} - \overline{2} - 2\big) = P\big(\overline{2}\big) \cdot P\big(\overline{2}\big) \cdot P\big(2\big) = {5\over6}\cdot {5\over6}\cdot{1\over6} = {25\over216}$$

Given that all 3 branches have the same above probability, the probability for all 3 branches is just 3 times as much:

$$P\big(\text{single 2 in 3 throws}\big) = 3 \cdot P\big(\overline{2} - \overline{2} - 2\big) = {75\over216} = {25\over72} $$

  • $\begingroup$ what answer does it give? $\endgroup$ – pi-π Aug 21 '16 at 16:40
  • $\begingroup$ Is it $\frac {1}{8}$ or $\frac {25}{72}$? $\endgroup$ – pi-π Aug 21 '16 at 16:41
  • $\begingroup$ @user354073 I edited my answer. I think the true problem is to understand the task in the textbook exactly. They you worded it here: "scoring a 2 on just one occasion" might be different from what the task reads in the book. Be sure to read it carefully and pay attention to the details. $\endgroup$ – null Aug 21 '16 at 17:08

Maybe don't draw all $216$ branches, just draw all the ones which contain a $2$ and combine it with your knowledge that the size of the sample space is $216$

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You can just count the number of outcomes which contain a $2$. If we go with the left branch after Roll $1$, we have "A $2$", and then in Roll $3$ we have $5$ choices that are "Not a $2$", a success. If we follow the right branch after Roll $1$, we have "$5$ branches" with "Not a $2$", and each has only one branch with a $2$, so now our total is up to $10$.

You can do this for each $1, 2, 3, 4, 5, 6$ in Roll $1$, and they will all be basically the same except when Roll $1$ is a $2$.

Now we have $10$ outcomes which contain only one $2$ for each of Roll $1 = 1, 3, 4, 5, 6$, so we have $50$. When Roll $1 = 2$, we have $25$ outcomes which contain only one $2$. So the correct answer should be $\dfrac {75}{216}$, or $\dfrac {25}{72}$

$$P(\text{exactly one 2}) = \dfrac {25}{72}$$


To help you with the tree, consider the following approach.

Rather than considering all the $216$ cases, just split on whether you get a $2$ at every single throw. Your tree should just consist of $2^3=8$ leaves.


a short answer but all other are already clear. There are 3 ways to get only one 2 once :

  • getting a 2 with the first throw and not a 2 with the 2 others : $\frac16 \frac56 \frac56$
  • getting a 2 with the 2nd throw and not a 2 with the 2 others : $\frac56 \frac16 \frac56$
  • and getting a 2 with the 3rd throw and not a 2 with the 2 others : $\frac56 \frac56 \frac16$

Then sum the probabilities of each case and you will find $3 \frac56 \frac16 \frac56 = 25/72$

Now the challenge is to find what modification of the question leads to a probability of $\frac18$, apart 3 times an even number.


The easiest way to do this is without a probability tree at all; although if you were to draw a tree diagram, you would be better off drawing branches for '2' and 'not 2' as oppose to '1', '2', '3', '4', '5' and '6' (the former gives you 2 branches to the power of 3 rolls = 8 branches, whereas the latter gives you 3^6 = 216 as you said).

The easiest solution is to use the binomial distribution. The binomial distribution calculates the probability of something happening a distinct number of times given the number of chances (rolls in this case) and the probability (1/6 in this case). To calculate this, consider the following expansion:


This binomial expansion represents every combination of $\frac{1}{6}$ and $\frac{5}{6}$ if you had to pick 3 altogether - the possibilities are (I will use '2' for rolling a 2, 'N' for rolling a not 2):

222, 22N, 2N2, N22, 2NN, N2N, NN2, NNN

If you count them you'll see one lot of 222, three of 22N (just in different orders), three of 2NN (again in different orders) and one lot of NNN - which correspond to the coefficients of the terms in the expansion above. So to calculate the probability of getting exactly one 2 and two 'not 2s' is P(2) * P(Not 2) * P(Not 2) * 3, which is what is written in the expansion above and computes to:


Now you might be thinking 'What about exactly 6 twos out of 10 rolls? I don't want to expand all that!'. The shortcut is Pascal's Triangle - or you can just use the formula:

$$X \text{~} Bin(10, \frac{1}{6})$$ $$P(X=6)=\binom{10}{6}(\frac{1}{6})^6(\frac{5}{6})^4=\frac{10!}{6!(10-6)!}(\frac{1}{6})^6(\frac{5}{6})^4=0.00217.....$$

That means 'X follows the binomial distribution with 10 tries and probability of success 1/6'. The second line means 'probability of getting exactly 6 successes is 10 Choose 2 (the 10 above the 2) x success^6 x failure^4'. The formula for n-choose-r (number of ways to choose exactly r elements out of n) is


  • $\begingroup$ @Mico Corrected, thanks. $\endgroup$ – otah007 Aug 22 '16 at 19:04

The 5 non 2 dies all have the same tree each one giving 10 paths out of 36 for a true answer which gives 50 /180 true paths . The 1st 2 die tree results in 25 out of 36 true paths which equal 75/216 in total or 25/72.


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