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#### chelseajjc95

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- Jun 23, 2018

- 4

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')

- Thread starter chelseajjc95
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- Thread starter
- #1

- Jun 23, 2018

- 4

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.

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')

- Jan 30, 2018

- 434

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

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')

(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

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