This is my second post . I still don't get and it is due tomorrow! !!
Jill made two 6 sided dice and labeled each od them with these numbers
-3, -2, -1, 0, 1, 2
Suppose this pair of dice was tossed 1000 times and the numbers on the two top faces were added. Which sum would probably be the most common? And why??
If -3 were one die, then the possible sums of two dies would be -6,-5,-4,-3,-2,-1: avg -3.5
if -2 were one die, then the possible sums of two dies would be -5,-4,-3,-2,-1,0: avg-2.5
the figure all possibilities for the first die, and the avg sum on each. Add all the sums, divide by six. That is the most probable sum.
In my head, I am seeing avg of sum of 1 for all combinations.
How about this:
create a table with 6 rows and 6 columns
label the rows: -3, -2, -1, 0, 1, 2
and the columns the same way
now fill in the entries by adding the row and column positions.
Look at the sums. Which is the most common and how many times does it happen?
multiply 1000 by that count over 36
In the table I suggested, the most common sum is -1, and it occurs 6 times.
(just like in 2 standard die, the most common sum is 7 and it occurs 6 times)
the prob of that most common sum is 6/36 = 1/6
If we were to toss our dice 1000 times, we could expect a sum of -1 appr
1000(1/6) or 167 times
To determine which sum would probably be the most common when the two dice are tossed 1000 times, we need to calculate the probability of each possible sum and compare them.
To do this, we'll list all the possible sums that can occur when rolling two 6-sided dice labeled with the numbers -3, -2, -1, 0, 1, and 2.
Let's start by listing all the possible outcomes when rolling the first die and their corresponding probabilities:
- Outcome: -3, Probability: 1/6
- Outcome: -2, Probability: 1/6
- Outcome: -1, Probability: 1/6
- Outcome: 0, Probability: 1/6
- Outcome: 1, Probability: 1/6
- Outcome: 2, Probability: 1/6
Similarly, we'll list all the possible outcomes for the second die and their corresponding probabilities:
- Outcome: -3, Probability: 1/6
- Outcome: -2, Probability: 1/6
- Outcome: -1, Probability: 1/6
- Outcome: 0, Probability: 1/6
- Outcome: 1, Probability: 1/6
- Outcome: 2, Probability: 1/6
Next, we'll calculate the probability for each possible sum by multiplying the probabilities of the corresponding outcomes. For example, the sum of -3 and -3 would have a probability of (1/6) * (1/6) = 1/36.
Here are all the possible sums and their corresponding probabilities:
- Sum: -6, Probability: 1/36
- Sum: -5, Probability: 2/36
- Sum: -4, Probability: 3/36
- Sum: -3, Probability: 4/36
- Sum: -2, Probability: 5/36
- Sum: -1, Probability: 6/36
- Sum: 0, Probability: 5/36
- Sum: 1, Probability: 4/36
- Sum: 2, Probability: 3/36
- Sum: 3, Probability: 2/36
- Sum: 4, Probability: 1/36
Now, to determine the most common sum, we look for the sum with the highest probability. From the list above, we can see that the sums -1 and 1 both have a probability of 6/36, which is the highest probability. Therefore, the sum of -1 and 1 would probably be the most common when the two dice are tossed 1000 times.
Explanation: The sum of -1 and 1 is the most common because it has the highest probability among all the possible sums. Since each face of the dice is equally likely to appear, the sums that are closer to zero (like -1 and 1) have a higher chance of occurring compared to the sums farther away from zero.