Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (anonymous):

assume that Pr[A ∪B] = 0.3 and Pr[A] = 0.2. (1) What is Pr[B] if A and B are independent events?

OpenStudy (anonymous):

Pr[A U B] = Pr[A] + Pr[B] - Pr[AB] That's the definition you need. You also need to know the definition of independent events. You should be able to solve it in one or two lines after that.

OpenStudy (anonymous):

im confused the disjoint is .1 so what is the independent?

OpenStudy (anonymous):

Independence implies Pr[AB] = Pr[A]*Pr[B] You may want to read your textbook. These definitions should be in there and are pretty important to solving these equations.

OpenStudy (anonymous):

how do u know what b is when they only give u a

OpenStudy (anonymous):

Also, disjoint implies Pr[AB] = 0. The intersection of two disjoint sets is empty. You're solving for Pr[B].

OpenStudy (anonymous):

i know u need to multiply the 2

OpenStudy (anonymous):

so what is b?

OpenStudy (anonymous):

I'm not giving you the answer. You have the equations you need. Plug in the numbers and solve it. I'll check your answer for you, if you'd like.

OpenStudy (anonymous):

i got .1

OpenStudy (anonymous):

that is the disjoint i need the independent

OpenStudy (anonymous):

Again, use the definition: A and B independent implies Pr[AB] = Pr[A]*Pr[B]

OpenStudy (anonymous):

would it be 0?

OpenStudy (anonymous):

Not quite. Here, I'll write the equation for you: P[A U B] = P[A] + P[B] - P[A]*P[B] You're solving for P[B]. You know P[A] and P[A U B]. Just plug in the numbers.

OpenStudy (anonymous):

i know its not working u can't do .1=x-.2x

OpenStudy (anonymous):

i figured it out thanks

OpenStudy (anonymous):

Good. :)

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!