MHB Find P(Q') B) P(Q union R) C) P(R')

[chelseajjc95]

I have this problem with understanding this specific notation could someone explain this notation & figure out how to solve this problem!?

Suppose P(Q)=5/31 , P(R)= 7/31 , P(Q intersect R)=3/31
Find the value of:
A) P(Q') B) P(Q union R) C) P(R')

 

[tkhunny]

I have this problem with understanding this specific notation could someone explain this notation & figure out how to solve this problem!?

Suppose P(Q)=5/31 , P(R)= 7/31 , P(Q intersect R)=3/31
Find the value of:
A) P(Q') B) P(Q union R) C) P(R')

I might be tempted to draw two partially intersecting circles. Call one Q and the other R. Draw a box around them.

 

[HOI]

I have this problem with understanding this specific notation could someone explain this notation & figure out how to solve this problem!?

Suppose P(Q)=5/31 , P(R)= 7/31 , P(Q intersect R)=3/31
Find the value of:
A) P(Q') B) P(Q union R) C) P(R')

If you really have no idea what the notations even mean where in the world did you get this question?

(A) Q' is the event that Q does NOT happen.
If the probability that Q happens is P(Q)= 5/31 then the probability that Q does not happen is P(Q')= 1- P(Q)= 1- 5/31= 26/31.

(B) Q union R is the event that either Q or R or both happen. Because it is possible that both Q and R happen (which is "Q intersect R"), if we just added P(Q) and P(R) we would have counted the times when both happen twice. We can fix that by subtracting off one P(Q intersect R). That is P(Q union R)= P(Q)+ P(R)- P(Q intersect R)= 5/31+ 7/31- 3/31= 9/31.

(C) As in (A), R' is the event that "R does NOT happen". P(R')= 1- P(R)= 1- 7/31= 24/31.