Answer:
a) The probability mass function of K = [tex]P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...[/tex]
b)
c)
Step-by-step explanation:
a) Let p be the probability of winning each ticket be = 0.1
Then q which is the probability of failing each ticket = 1 - p = 1 - 0.1 = 0.9
Assume X represents the number of failure preceding the 5th success in x + 5 trials.
The last trial must be success whose probability is p = 0.1 and in the remaining (x + r- 1) ( x+ 4 ) trials we must have have (4) successes whose probability is given by:
[tex]\binom{x+r-1}{r-1}*p^{r-1}*q^{x} = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........[/tex]
Then, the probability distribution of random variable X is
[tex]P(X=x) = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........[/tex]
where;
X represents the negative binomial random variable.
K= X + 5 = number of ticket buy up to and including fifth winning ticket.
Since K =X+5 this signifies that X = K-5
as X takes value 0, 1 ,2,...
K takes value 5, 6 ,...
Therefore:
The probability mass function of K = [tex]P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...[/tex]
b)
Let p represent the probability of getting a tail on a flip of the coin
Thus p = 0.5 since it is a fair coin
where L = number of flips of the coin including 33rd occurrence of tails
Thus; the negative binomial distribution of L can be illustrated as:
[tex]P(X=x) = \binom{x-1}{r-1}(1-p)^{x-r}p^r[/tex]
where
X= L
r = 33 &
p = 0.5
Since we are looking at the 33rd success; L is likely to be : L = 33,34,35...
Thus; the PMF of L = [tex]P(L=l) = \binom{l-1}{33-1}(1-0.5)^{l}(0.5)^{33} \\ \\ \\ \mathbf{P(L=l) = \binom{l-1}{33-1}(0.5)^{l} }[/tex]
c)
Given that:
Let M be the random variable which represents the number of tickets need to be bought to get the first success,
also success probability is 0.01.
Therefore, M ~ Geo(0.01).
Thus, the PMF of M is given by:
[tex]P(M = m) = (1-0.01)^{m-1} * 0.01 , \ \ \ since \ \ \ (m = 1,2,3,4,....)[/tex]
[tex]P(M=m) = (0.99)^{m-1} * 0.01 , m = 1,2,3,4,....[/tex]
