# Given the following: P(A) = 0.3 P(B) = 0.2 P(A or B) = 0.5 P(A and B) = 0 Which of the following is true? A and B are disjoint. A and B are neither disjoint nor independent. A and B are independent. A and B are disjoint and independent. Question
Probability and combinatorics Given the following: P(A) = 0.3 P(B) = 0.2 P(A or B) = 0.5 P(A and B) = 0
Which of the following is true?
A and B are disjoint.
A and B are neither disjoint nor independent.
A and B are independent.
A and B are disjoint and independent. 2021-01-09
Given: P(A) = 0.3
P(B) = 0.2
P(A or B) = 0.5
P(A and B) = 0
Two events are disjoint, if the events cannot occur at the same time. Since P(A and B) = 0, the events A and B cannot occur at the same time and thus events A and B are disjoint.
Two events are independent, if the probability that one event occur in no way affects the probability of the other event ocuring.
Moreover, two events are intependent if and only if P(A and B) = P(A) * P(B).
P(A) * P(B) = 0.3 * 02 = 0.06
P(A and B) = 0
Since 0.06 ≠ 0, the events A and B are not independent. Thus the answer is then: A and B are disjoint.

### Relevant Questions A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet $1 onred. If red appears (which has probablity 18/38), then the gamblershould take her$1 profit and quit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional \$1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gambler's winnings when she quites.
(a) Find P{X > 0}.
(b) Are you concinved that the strategy is indeed a "winning" strategy? Explain your answer.
(c) Find E[X]. A radio station gives a pair of concert tickets to the 6th called who knows the birthday of the performer. For each person who calls, the probability is.75 of knowing the performer birthday. All calls are independent.
a. What is the PMF (Probability Mass Function) of L, the number of calls necessary to find the winner?
b. What is the probability of finding the winner on the 10th call?
c. What is the probability that the station will need 9 or more calls to find a winner? A radio station gives a pair of concert tickets to the six caller who knows the birthday of the performer. For each person who calls, the probability is 0.75 of knowing the performer's birthday. All calls are independent.
a) What is the PMF of L, the numberof calls necessary to find the winner? MSK b) What is the probability of finding the winner on the tenth caller?
c) What is the probability of finding the winner on the tenth caller? Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}. Which of the following is(are) True?
I. The means of the Student’s t and standard normal distributions are equal.
II. The standard normal distribution approaches the Student’s t distribution as the degrees of freedom becomes large.
A. I only
B. II only
C. Both I and II
D. Neither I nor II In one study, the correlation between the educational level of husbands and wives in a certain town was about 0.50, both averaged 12 years of schooling completed, with an SD of 3 years.
a) Predict the educational level of a woman whose husband has completed 18 years of schooling b) Predict the educational level of a man whose wife has completed 15 years of schooling. c) Apparently, well-educated men marry women who are less well educated than themselves. But the women marry men with even less education. How is this possible? A two-sample inference deals with dependent and independent inferences. In a two-sample hypothesis testing problem, underlying parameters of two different populations are compared. In a longitudinal (or follow-up) study, the same group of people is followed over time. Two samples are said to be paired when each data point in the first sample is matched and related to a unique data point in the second sample.
This problem demonstrates inference from two dependent (follow-up) samples using the data from the hypothetical study of new cases of tuberculosis (TB) before and after the vaccination was done in several geographical areas in a country in sub-Saharan Africa. Conclusion about the null hypothesis is to note the difference between samples.
The problem that demonstrates inference from two dependent samples uses hypothetical data from the TB vaccinations and the number of new cases before and after vaccination. PSK\begin{array}{|c|c|} \hline Geographical\ regions & Before\ vaccination & After\ vaccination\\ \hline 1 & 85 & 11\\ \hline 2 & 77 & 5\\ \hline 3 & 110 & 14\\ \hline 4 & 65 & 12\\ \hline 5 & 81 & 10\\\hline 6 & 70 & 7\\ \hline 7 & 74 & 8\\ \hline 8 & 84 & 11\\ \hline 9 & 90 & 9\\ \hline 10 & 95 & 8\\ \hline \end{array}ZSK
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.  The problem reads: Suppose $$\displaystyle{P}{\left({X}_{{1}}\right)}={.75}$$ and $$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}={.40}$$. What is the joint probability of $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$?
This is how I answered it. P($$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$) $$\displaystyle={P}{\left({X}_{{1}}\right)}\times{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}={.75}\times{.40}={0.3}.$$
What I don't understand is how do you get the $$\displaystyle{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}$$? I am totally new to Statistices and I need to understand each part of the process in order to get the whole concept. Can anyone help me to understand why the P and X exist and what they represent? Let the sequence of events E1, E2, . . . , En be independent, and assume that $$\displaystyle{P}{\left({E}{i}\right)}=\frac{{1}}{{{i}+{1}}}$$. Show that $$\displaystyle{P}{\left({E}{1}∪···∪{E}{n}\right)}=\frac{{n}}{{{n}+{1}}}$$