A die is rolled 5 times, what is the probability that there are exactly 2 fives? A die is rolled 5 times. 
Determine the probability that there are exactly 2 fives. The answer should be a decimal.
 A: Let's look at one possible way:
$$x_1, 5, x_2, 5, x_3$$
The $x_i$ stand for "anything else but 5." The probability of this event is $(\frac{1}{6})^2(\frac{5}{6})^3$. So, it can happen in this particular way OR (this is the key word) another way corresponding to all the other ways to arrange the fives. How many ways are there to arrange two objects in a set of five?
The answer is $\binom {5}{3}$. Hence, the probability of your event is $$\binom {5}{3}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3.$$
A: The total number of combinations is $6^5$. To get a combination with exactly 2 fives, there are $1\times1\times5\times5\times5=5^3$ ways of arranging the dice if the first two dice are fives. Multiply that by the number of ways of arranging the two fives amongst the 5 dice - $\binom{5}{2}=\frac{5!}{2!\cdot3!}=10$. Thus, $$\binom{5}{2}\cdot\frac{5^3}{6^5}\approx0.161$$ Is the proportion of throws that will have exactly 2 fives.
A: If you require two specific times(e.g. first time and second time) to be 5(probability $1/6$), others to be non-5(probability $1-1/6=5/6$). The probabiliy is $(1/6)^2(5/6)^3$, you have $C^5_2=10$ different ways to roll two 5. Therefore the answer is $10(1/6)^2(5/6)^3$, which is approximately 0.1608.
