A fair die is rolled repeatedly. Calculate to at least two decimal places: a) the chance that the first 6 appears before the tenth roll
b) the chance that the third 6 appears on the tenth roll
c) the chance of seeing three 6's among the first ten rolls given that there were six 6's among the first twenty roles.
d) the expected number of rolls until six 6's appear
e) the expected number of rolls until all six faces appear

a. 0.34885

b. 0.04651

c. 0.02404

d. 36

e. 14.7, say 15 trials

Step-by-step explanation:

Q17070205

Note:  

1. In order to be applicable to established probability distributions, each roll is considered a Bernouilli trial, i.e. has only two outcomes, success or failure, and are all independent of each other.

2. use R to find the probability values from the respective distributions.

a) the chance that the first 6 appears before the tenth roll

This means that a six appears exactly once between the first and the nineth roll.

Using binomial distribution, p=1/6, n=9, x=1

dbinom(1,9,1/6) = 0.34885

b) the chance that the third 6 appears on the tenth roll

This means exactly two six's appear between the first and 9th rolls, and the tenth roll is a six.

Again, we have a binomial distribution of p=1/6, n=9, x=2

p1 = dbinom(2,9,1/6) = 0.27908

The probability of the tenth roll being a 6 is, evidently, p2 = 1/6.

Thus the probability of both happening, by the multiplication rule, assuming independence  

P(third on the tenth roll) = p1*p2 = 0.04651

c) the chance of seeing three 6's among the first ten rolls given that there were six 6's among the first twenty roles.

Again, using binomial distribution, probability of 3-6's in the first 10 rolls,

p1 = dbinom(3,10,1/6) = 0.15504

Probability of 3-6's in the NEXT 10 rolls

p1 = dbinom(3,10,1/6) = 0.15504

Probability of both happening  (multiplication rule, assuming both events are independent)

= p1 *  p1 = 0.02404

d) the expected number of rolls until six 6's appear

Using the negative binomial distribution, the expected number of failures before n=6 successes, with probability p = 1/6

=  n(1-p)/p

Total number of rolls by adding n  

= n(1-p)/p + n = n(1-p+p)/p = n/p = 6/(1/6) = 36

e) the expected number of rolls until all six faces appear

P1 = 6/6 because the firs trial (roll) can be any face with probability 1

P2 = 6/5 because the second trial for a different face has probability 5/6, so requires 6/5 trials

P3 = 6/4 ...

P4 = 6/3

P5 = 6/2

P6 = 6/1

So the total mean (expected) number of trials is 6/6+6/5+6/4+6/3+6/2+6/1 = 14.7, say 15 trials

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