Answer to Question #128080 in Statistics and Probability for ysabe castro

Statistics and Probability

Question #128080

Given P(E) = 0.25, P(F) = 0.6, and P(E ∪ F) = 0.7.
Find:

Expert's answer

P(E\cup F)=P(E)+P(F)-P(E\cap F)

P(E\cap F)=P(E)+P(F)-P(E\cup F)=

=0.25+0.6-0.7=0.15

Since $P(E\cap F)\not =0,$ then event $E$ and event $F$ are not mutually exclusive.

P(E)\cdot P(F)=0.25\cdot0.6=0.15=P(E\cap F)

Event $E$ and event $F$ are independent.

Learn more about our help with Assignments: Statistics and Probability

Comments

andy

15.03.21, 18:32

Event E and event F are independent

Leave a comment

Related Questions

1. a. What is the probability of the first selected person is a left-handed men and the second s
2. The number of visits to a website in one hour has the following probability mass function{1/10 , �
3. a) A box contains 25 Balls, 20% of them are white whereas the rest of them are black. A person rando
4. The thickness of books in a bookstore follows a normal distribution with the mean of 45mm and the st
5. In a school, 60% of the students take a foreign language class and 20% of students take both foreign
6. (a) Let A = {2,4,6} and B = {x,y,z}. State each of the following are relations from A into B. (i) R1
7. suppose x is a normally distributed random variable with μ = 50 and ϭ = 3. find a value of the ran