If 2 dice are tossed 1000 times, predict the approximate number of times each sum below would occur in 1000 tosses.
2:
3
4 and so on until 12
HOW DO U DO THIS?!
Each toss of two dice has 36 possibilites.
2 can only be obtained by 1,1 --> 1/36 *1000 = ?
4 can be obtained by
2,2
3,1
1,3
so --> 3/36 * 1000 = ?
And so on….
To predict the approximate number of times each sum would occur when two dice are tossed 1000 times, we can use the concept of probability.
First, let's understand the total number of possible outcomes when two dice are tossed. Each die has 6 faces, so the total number of outcomes is 6 multiplied by 6, which equals 36.
Next, we need to determine the number of ways each sum can be obtained. For example, the sum of 2 can only be obtained when both dice show a 1. There is only 1 way to get a sum of 2.
For the sum of 3, there are two possible combinations: (1, 2) and (2, 1). So, there are 2 ways to get a sum of 3.
For the sum of 4, the possible combinations are (1, 3), (2, 2), and (3, 1). So, there are 3 ways to get a sum of 4.
Similarly, you can calculate the number of ways for each sum until 12.
Now, divide the number of ways to obtain each sum by the total number of outcomes. This will give you the probability of getting each sum.
Finally, multiply the probability by the total number of tosses (1000) to get the approximate number of times each sum would occur in 1000 tosses.
Here's an example to clarify the process:
To calculate the expected number of times the sum of 7 would occur:
- The number of ways to obtain a sum of 7 is 6 (sums (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)).
- The total number of outcomes is 36 (6 faces on each die, so 6 x 6 = 36).
- The probability of getting a sum of 7 is 6/36 = 1/6.
- Finally, multiply the probability (1/6) by the total number of tosses (1000). The approximate number of times the sum of 7 would occur is 1000/6 ≈ 166.
Repeat this process for each sum from 2 to 12, and you can predict the approximate number of times each sum would occur in 1000 tosses.