After three tosses of a fair die, which is more likely to come up: three distinct digits (e.g. 1,2,3), or three digits with at least one repeat (e.g. 1,2,1)?
Distinct digits is more likely
At least one repeat is more likely
They are the same
Cannot determine
Explain
the only combinations with 3 distinct digits is
1 2 3 , 2 3 4 , 3 4 5 , and 4 5 6
Each of those can be arranged in 6 ways, so there are only 24 cases with the numbers distinct
the 3 dice can fall in 6^3 or 216 different ways, so in 192 of those ways, one or more digits would repeat.
So what do you think?
what about the 3 distinct digits of 124, 125, 126, 134, 135, 136, 145, 146, 156, 235, 236, 245, 246, 256, 346 & 356?
To determine which outcome is more likely, we need to calculate the probabilities of each event occurring.
Let's begin with the event of getting three distinct digits. In order to have three distinct digits, the first toss of the die can be any number from 1 to 6. For the second toss, there are five remaining numbers that can be chosen (since we want the digits to be distinct from the first toss), and for the third toss, there are four remaining numbers that can be chosen. Therefore, the probability of getting three distinct digits is (1/6) * (5/6) * (4/6) = 20/216.
Now, let's consider the event of getting three digits with at least one repeat. This means that at least two of the three numbers must be the same. There are two cases to consider: either two numbers are the same, or all three numbers are the same.
For the first case, in order to have two numbers that are the same, the first toss can be any number from 1 to 6. The second toss must then match the first toss, resulting in a probability of 1/6. For the third toss, there are five remaining numbers that can be chosen (since we want it to be different from the first two tosses). Therefore, the probability of getting two numbers the same is (1/6) * (1/6) * (5/6) = 5/216.
For the second case, in order to have all three numbers the same, the first toss can be any number from 1 to 6. The second toss must then also be the same as the first toss, resulting in a probability of 1/6. Finally, the third toss must also be the same as the first two tosses, resulting in a probability of 1/6. Therefore, the probability of getting all three numbers the same is (1/6) * (1/6) * (1/6) = 1/216.
To find the total probability of getting three digits with at least one repeat, we can add the probabilities from the two cases: (5/216) + (1/216) = 6/216.
Comparing the probabilities, we can see that 20/216 is greater than 6/216. Therefore, the event of getting three distinct digits (e.g., 1, 2, 3) is more likely than getting three digits with at least one repeat (e.g., 1, 2, 1). Thus, the answer is "Distinct digits is more likely."