The probability of scoring a 2 just once when a die is thrown three times? 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.
 A: 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 $216$

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


*

*going along a branch means multiplying probabilities and the result is the probability of the entire branch to happen

*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:
$$P\big(\text{single 2 in 3 throws}\big) = 3 \cdot P\big(\overline{2} - \overline{2} - 2\big) = {75\over216} = {25\over72} $$
A: 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: 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:
$$(\frac{1}{6}+\frac{5}{6})^3=(\frac{1}{6})^3+3(\frac{1}{6})^2(\frac{5}{6})+3(\frac{1}{6})(\frac{5}{6})^2+(\frac{5}{6})^3$$
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:
$$3(\frac{1}{6})(\frac{5}{6})^2=\frac{25}{72}$$
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
$$\frac{n!}{r!(n-r)!}$$
A: 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.
A: 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.
