Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

What are dependent events? and an example?

OpenStudy (amistre64):

dependant events happen as a result of one causing the other; there is some causal relation between them right?

OpenStudy (anonymous):

example: pick a ball from an urn containing red and what balls and then WITHOUT REPLACING IT pick another. the probability of picking a red ball on the second try will change depending on whether or not the first all selected was red

OpenStudy (anonymous):

simpler example. roll two dice. if the first one shows "1" then the probability that the total is 8 given the first one is "1" is zero. whereas if the first one is 2, then the probability that the total is 8 given the first one is 2 is \[\frac{1}{6}\]

OpenStudy (anonymous):

DEPENDENT EVENTS In some cases one event is dependent on another; that is, two or more events are said to be dependent if the occurrence or nonoc­currence of one of the events affects the probabilities of occur­rence of any of the others. Consider that two or more events are dependent. If p1 is the probability of a first event; p2 the probability that after the first happens, the second will occur; p3 the probability that after the first and second have happened, the third will occur; etc., then the probability that all events will happen in the given order is the product p1*p2*p3.

OpenStudy (anonymous):

in fact although we make frequently make an assumption of independence in general it is rare

OpenStudy (anonymous):

Thank You!

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

btw two events A and B are independent if \[P(A|B)=P(A)\]meaning knowing whether or not B has occurred gives us no additional knowledge about the probability of A occurring

OpenStudy (anonymous):

Exampl: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack? Probabilities: P(queen on first pick) = 4/52 P(jack on 2nd pick given queen on 1st pick) = 4/51 P(queen and jack) = 4/52 · 4/51 = 16 /2652 = 4/663

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!