Q1: A Pair Of Dice Tossed Repeatedly What Is The Least Numbe

Q1 A Pair Of Dice Is Tossed Repeatedly What Is The Least Number O

A pair of dice is tossed repeatedly. What is the least number of tosses needed so that the probability of getting a total of 11 exceeds certain thresholds?

Paper For Above instruction

Introduction

The problem involves understanding the probability of rolling a sum of 11 with two six-sided dice and determining the minimum number of such rolls so that this probability exceeds specified values, namely 0.5 and 0.95. This type of problem exemplifies the application of binomial probability and the law of large numbers in probability theory. The focus is on leveraging the probability of the event in a single trial, computing the cumulative probability over multiple independent trials, and solving inequalities to find the least number of trials satisfying the conditions.

Probability of Rolling a Sum of 11 with Two Dice

There are a total of 36 possible outcomes when two standard dice are rolled (6 faces on each die). The specific outcomes that sum to 11 are: (5,6) and (6,5). There are 2 favorable outcomes, so the probability of rolling a sum of 11 in one throw is:

P(11) = 2/36 = 1/18 ≈ 0.0556.

Part (i): Probability exceeds 0.5

Let n be the number of independent tosses needed. The probability of not obtaining a total of 11 in a single toss is:

P(not 11) = 1 - 1/18 = 17/18 ≈ 0.9444.

The probability that none of the n tosses result in a sum of 11 is:

P(no 11 in n tosses) = (17/18)^n.

Therefore, the probability that at least one sum of 11 occurs in n tosses is:

P(at least one 11 in n tosses) = 1 - (17/18)^n.

We want this to be greater than 0.5:

1 - (17/18)^n > 0.5

which simplifies to:

(17/18)^n

Taking logarithms:

n ln(17/18)

Note that ln(17/18) is negative, so inequality reverses:

n > ln(0.5) / ln(17/18)

Calculating:

- ln(0.5) ≈ -0.6931

- ln(17/18) ≈ -0.0583

Thus:

n > -0.6931 / -0.0583 ≈ 11.89

The least integer satisfying this is n = 12.

Part (ii): Probability exceeds 0.95

Similarly, for 0.95:

(17/18)^n

Taking logarithms:

n ln(17/18)

Calculating:

- ln(0.05) ≈ -2.9957

Thus:

n > -2.9957 / -0.0583 ≈ 51.37

The least integer is n = 52.

Conclusion

The minimum number of tosses needed so that the probability of obtaining at least one sum of 11 exceeds:

- 0.5 is 12,

- 0.95 is 52.

References

  • Ross, S. M. (2014). Introduction to Probability Models (11th Ed.). Academic Press.
  • Gut, A. (2013). An Intermediate Course in Probability. Springer.
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  • Larson, R., & Farrier, R. (2018). Elementary Statistics. Cengage Learning.
  • Kimmel, M., & Axelrod, D. E. (2015). Chance Processes: A First Course in Probability. Springer.
  • Grinstead, C. M., & Snell, J. L. (1997). Introduction to Probability. American Mathematical Society.
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