Question

One hundred random draws with replacement are made from the box [1, 6, 7, 9, 9, 10] (a) Calculate the probability of getting 12 nines. (b) Calculate the probability that the sum is at most 107.

Answer 1

a) Probability of getting a nine in any draw is computed as:
= Total number of 9s in the lot / Total number of numbers
= 2/6
= 1/3

Therefore the number of 9s that can be obtained from the 100 draws could be modelled here as a binomial distribution given as:

Probability of getting 12 nines is computed using binomial probability function here as:

Therefore 0 is the approx. probability here.

b) The expected value here for any draw is computed as:

E(X) = (1 + 6 + 7 + 9 + 9 + 10)/6 = 42/6 = 7

The second moment is computed as:
E(X²) = (1² + 6² + 7² + 9² + 9² + 10²)/6 = 58
Therefore Var(X) = E(X²) - [E(X)]² = 58 - 7² = 9

Therefore SD(X) = square root of Var(X) = square root of 9 = 3

Therefore the mean sample size of 100 drawn could be modelled here by central limit theorem as:

The probability that the sum is at most 107 is computed here as:

Note that sums of 107 or less could be obtained only in following ways:

• 100*1, Prob = (1/6)¹⁰⁰
• 99*1 + 6 = 105, Prob = 100*(1/6)¹⁰⁰
• 99*1 + 7 = 106, Prob = 100*(1/6)¹⁰⁰

Therefore, total probability here is computed as:
= 201*(1/6)¹⁰⁰ which is again approximately equal to 0.