Showing Dependence of Events in Probability Theory, Assignments of Probability and Statistics

An example of calculating the probability of intersections of events in probability theory using the given sample space s = {1, 2, 3, 4} and associated probabilities p({1}), p({2}), p({3}), and p({4}). The document also shows that the events e1 = {1, 3}, e2 = {2, 3}, and e3 = {3, 4} are not independent by demonstrating that their intersection's probability does not equal the product of their individual probabilities.

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

koofers-user-gqr
koofers-user-gqr 🇺🇸

10 documents

1 / 1

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Suppose S={1,2,3,4}and
P({1}) = 2
21
4
P({2}) = 1
4
P({3}) = 2
2+3
4
P({4}) = 1
4
E1={1,3}
E2={2,3}
E3={3,4}.
Show that: P(E1E2E3) = P(E1)P(E2)P(E3) but that no pair of events E1,E2, and E3are
independent and hence E1,E2, and E3are not independent.
1

Partial preview of the text

Download Showing Dependence of Events in Probability Theory and more Assignments Probability and Statistics in PDF only on Docsity!

Suppose S = { 1 , 2 , 3 , 4 } and

P ({ 1 }) =

P ({ 2 }) =^1

P ({ 3 }) = −

P ({ 4 }) =

E 1 = { 1 , 3 }

E 2 = { 2 , 3 }

E 3 = { 3 , 4 }.

Show that: P (E 1 ∩ E 2 ∩ E 3 ) = P (E 1 )P (E 2 )P (E 3 ) but that no pair of events E 1 ,E 2 , and E 3 are independent and hence E 1 ,E 2 , and E 3 are not independent.