1. When two fair dice are tossed, what is the probability of a sum of 4 or a sum of 8?
2. When two fair dice are tossed, what is the probability of doubles or a sum over 9?
3. Evaluate 1025! / 1022!
Please help
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There are 36 possibilities with 2 dice.
sum of 4 = 2,2 or 1,3 or 3,1
Sum of 8 = 2,6 or 6,2 or 3,5 or 5,3 or 4,4
1. 8/36
2. Use similar process.
3. 1023 * 1024 * 1025 = ?
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1. To determine the probability of the sum of 4 or 8 when tossing two fair dice, we need to find the number of favorable outcomes and divide it by the total number of possible outcomes.
To find the number of favorable outcomes for a sum of 4, we have the combinations (1,3) and (3,1), which equals 2. For a sum of 8, we have combinations (2,6), (6,2), (3,5), (5,3), and (4,4), which equals 5. Therefore, the total number of favorable outcomes is 2 + 5 = 7.
The total number of possible outcomes is 6 * 6 = 36, since each dice can have 6 possible outcomes (numbers 1 to 6).
So, the probability of a sum of 4 or 8 is 7/36.
2. To calculate the probability of rolling doubles or a sum over 9 with two fair dice, we again need to find the number of favorable outcomes and divide by the total number of possible outcomes.
For doubles, we have the combinations (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6), which equals 6.
For a sum over 9, we have the combinations (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6), which equals 6.
So, the total number of favorable outcomes is 6 + 6 = 12.
The total number of possible outcomes remains the same as before, 6 * 6 = 36.
Therefore, the probability of rolling doubles or a sum over 9 is 12/36, which simplifies to 1/3.
3. To evaluate 1025!/1022!, we can simplify it by canceling out common factors between the numerator and the denominator.
We know that n!/m! = n*(n-1)*(n-2)*...*(m+1), where n! represents the factorial of n.
In this case, we have 1025!/1022! = 1025 * 1024 * 1023.
Therefore, the value of 1025!/1022! is 1025 * 1024 * 1023. You can calculate this by multiplying the three numbers together to get the final result.