We throw a fair die 3 times.Find the probability to bring two times the same number

Question

We throw a die(fair) $3$ times.Find the probability to have two times the same number.

solution:i think $$\left(\frac{1}{6}\right)^3\cdot 6=\frac{1}{36}$$ Is this correct or am I missing something?

Answer 1

Assume we write the results of the throws on a paper, then here's how you can calculate all good combinations. Choose a random number (6 choices) and then write that number in two of places $(\binom{3}{2} = 3$ choices) And for the remaining number you can choose any of the rest $5$ numbers. Hence:

$$\text{Good Combinations} = 6 \cdot 3 \cdot 5 = 90$$

Divide this by the number of all possible combinations and you will get the final probability.

The previous method is to find the probability if you want the same number appearing exactly twice. If a good combination means that a number is appearing at least twice then find all bad combinations ($6 \cdot 5 \cdot 4 = 120$ choices) and subtract that number from all the combinations to get the number of good combinations and eventually find the probability of that happening.

Answer 2

1) Find the probability to have EXACTLY two times the same number: there are 6 ways to choose the value of the couple, 3 ways to assign them to the 3 dice (xxo,xox,oxx), and 5 ways to choose the third value (different form the couple). Hence $$p=\frac{6\cdot 3 \cdot 5 }{6^3}=\frac{15}{36}=\frac{5}{12}.$$

2) Find the probability to have AT LEAST two times the same number: just add to the previous cases the 6 triples with the same value. Hence $$p=\frac{6\cdot 3 \cdot 5 +6}{6^3}=\frac{15+1}{36}=\frac{4}{9}.$$