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Probability Sum of Two Dice Video Lecture | Engineering Mathematics - Civil Engineering (CE)

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FAQs on Probability Sum of Two Dice Video Lecture - Engineering Mathematics - Civil Engineering (CE)

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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