Mathematics, 11.06.2020 23:57 vegeta8375

Problem 3.3.9 • (a) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner with probability 0.1. Find the PMF of K, the number of tickets you buy up to and including your fifth winning ticket. (b) L is the number of flips of a fair coin up to and including the 33rd occurrence of tails. What is the PMF of L? (c) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner with probability 0.01. Let M equal the number of tickets you buy up to and including your first winning ticket. What is the PMF of M?

Answers: 1

Show answers

Answers

Answer from: Mikey5329

a) The probability mass function of K = $P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...$

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:

$\binom{x+r-1}{r-1}*p^{r-1}*q^{x} = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........$

Then, the probability distribution of random variable X is

$P(X=x) = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........$

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 = $P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...$

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:

$P(X=x) = \binom{x-1}{r-1}(1-p)^{x-r}p^r$

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 = $P(L=l) = \binom{l-1}{33-1}(1-0.5)^{l}(0.5)^{33} \\ \\ \\ \mathbf{P(L=l) = \binom{l-1}{33-1}(0.5)^{l} }$

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:

$P(M = m) = (1-0.01)^{m-1} * 0.01 , \ \ \ since \ \ \ (m = 1,2,3,4,....)$

$P(M=m) = (0.99)^{m-1} * 0.01 , m = 1,2,3,4,....$

Answer from: Quest

answer: c i think

step-by-step explanation:

The owners of four companies competing for a contract are shown in the table below

Answer from: Quest

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range:

Another question on Mathematics

Mathematics, 21.06.2019 14:00

The revenue generated by a bakery over x months, in thousands of dollars, is given by the function f(x) = 2(1.2)* the cost of running the bakery forx months, in thousands of dollars, is given by the function g(x) = 2x + 1.4determine the equation for h if h(x) = f(x) - g(x).oa. m(x) = (1-2)*-x-07b.(x) = 2(1 2 - 2x -0.7)h(x) = -2((1.2) + x + 0.7)d.h(x) = 2((12) - x-0.7)

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

Which of the following equations is of the parabola whose vertex is at (2, 3), axis of symmetry parallel to the y-axis and p = 4? a.)y-3 = 1/16 (x-2)^2 b.)y+3 = -1/16 (x+2)^2 c.)x-2 = 1/16 (y-3)^2

Answers: 3

Answer

Mathematics, 21.06.2019 17:30

What number should be added to the expression x^2+3x+ in order to create a perfect square trinomial? 3/2 3 9/4 9

Answers: 1

Answer

Mathematics, 21.06.2019 18:30

Find the constant of variation for the relation and use it to write an equation for the statement. then solve the equation.

Answers: 1

Answer

You know the right answer?

Problem 3.3.9 • (a) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner...