Answer:
a. 0.34885
b. 0.04651
c. 0.02404
d. 36
e. 14.7, say 15 trials
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