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Probability Sum of Two Dice Video Lecture | Engineering Mathematics - Engineering Mathematics
FAQs on Probability Sum of Two Dice Video Lecture - Engineering Mathematics - Engineering Mathematics
| 1. What is the probability of rolling a sum of 7 with two dice? |  |
Ans. The probability of rolling a sum of 7 with two dice is 1/6. This is because there are 6 possible outcomes that yield a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Since there are a total of 36 possible outcomes when rolling two dice (6 outcomes for the first die and 6 outcomes for the second die), the probability is 6/36, which simplifies to 1/6.
| 2. What is the probability of rolling a sum of 2 with two dice? | |
Ans. The probability of rolling a sum of 2 with two dice is 1/36. This is because there is only one outcome that yields a sum of 2: (1, 1). Since there are 36 possible outcomes when rolling two dice, the probability is 1/36.
| 3. How many different sums can be obtained when rolling two dice? | |
Ans. When rolling two dice, there are 11 different sums that can be obtained: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. These sums are obtained by adding up the numbers rolled on each die. For example, a sum of 7 can be obtained by rolling a (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1).
| 4. What is the most likely sum to be rolled with two dice? | |
Ans. The most likely sum to be rolled with two dice is 7. This is because there are more ways to obtain a sum of 7 than any other sum. As mentioned earlier, there are 6 possible outcomes that yield a sum of 7. In comparison, there are only 1 possible outcome for a sum of 2 or 12, and 2 possible outcomes for sums of 3, 4, 5, 11, and 10. Therefore, the probability of rolling a sum of 7 is the highest.
| 5. What is the probability of rolling a sum greater than 9 with two dice? | |
Ans. The probability of rolling a sum greater than 9 with two dice is 10/36, which simplifies to 5/18. This can be calculated by counting the number of outcomes that yield a sum greater than 9 (10 outcomes: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), (3, 7), (4, 8), (5, 7), and (5, 8)) and dividing it by the total number of possible outcomes (36 outcomes).
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