Math question part 2?

look for "math question" it has the information you need to answer the question.





I will give you the table you need though


men / women / total


less then one month's income 66 83 149


one month's income or more 76 62 138





TOTAL 142 145 287


(B) given that a randomly selected worker is male, find the the probability that the worker has less then one month's income


(C) given that a randomly selected worker has one month's income or more, find the probability that the worker is female.


(D) Are the events of having less than one month's income or more saved and being male independant or dependant explain.





PLEASE SHOW WORK I REALLY WANT TO UNDERSTAND HOW TO DO this i have talked with my teacher but i am still lost and there are no tutors aviable so please help me. thanks.

Math question part 2?
So the way I see it, the simplest way to understand this is through a probability table.





x1 = man


x2 = woman





y1 = less than one month's income


y2 = one month's income or more





imagine a 2x2 table:


x1, y1 || x2, y1


x1, y2 || x2, y2





x1, y1 means a male (x1) who has less than one month's income (y1). Likewise, x1, y2 means a male (x1) who has more than one month's income or more... and so forth for x2, y1 and x2,y2





You can also substitute the # of people into the table, so


66 || 83


76 || 62





now instead of using those individual #s, lets transform them to probabilities.





You know the total # of men is 66+76 = 142


You know the total # of women is 83+62 = 145


You know the total # of people based on gender is 287


You know the total # of people with less than one month's income is 149


You know that the total # of people with one month's income or more is 138.


You know the total of people using their income is 287.





The total # of people based on gender and the total # of people based on income should ALWAYS be the same.





So now, find the probability of x1,y1 or P(x1,y1). Since the # of people for x1,y1 is 66, you divide it by the total # of people, 287. So P(x1,y1) = 66/287. Likewise, for x1,y2 or P(x1,y2), you would do the same so it P(x1,y2) = 76/287





Here is the completed probability table


66/287 || 83/287


76/287 || 62/287





You can also convert the table in decimal form if you find that easier. The total probability for all possible outcomes should always be equal to 1.





Answering your Questions:





(B) Given that a randomly selected worker is male, find the the probability that the worker has less then one month's income.





You want to find P(worker has less than one month's income | worker is male). This is a conditional probability which means that you are given the condition that the worker is male, and out of those males, you want to know how many have less than one's month's income.





The formula used is P(A|B) = P(A^B) / P(B)





The total # of males in probability form is 66/287 + 76/287 = 142/287.





The probability of a male who has less than one month's income is 66/287.





So P(worker has less than one month's income | worker is male) = (66/287) / (142/287) or .464





(C) Given that a randomly selected worker has one month's income or more, find the probability that the worker is female.





Using the same formula P(A|B) = P(A^B) / P(B)





You know P(worker haso ne month's income or more) = 76/287 +62/287 = 138/287


You know P(female %26amp; has one month's income or more)= 62/287





So P(A|B) = (62/287) / (138/287) = .449





(D) Are the events of having less than one month's income or more saved and being male independant or dependant explain.





So the general formula for this is if P(A) x P(B) = P(A^B), then the events A and B are independent.


A=having less than one month's income


P(A) = 149/287


B=male


P(B) = 142/287


A^B=having less than one month's income %26amp; male


P(A^B) = 66/287





P(A) x P(B) = .2567 does not equal P(A^B) = .2299,


so A and B are not independent.

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