SOLUTION: Let A and B be events with P(A)=0.7 and P(B)=0.4, and P( A or B)= 0.8. (a) Compute P ( A and B). (b) Are A and B mutually exclusive? Explain (c) Are A and B independent?

Question 1148078: Let A and B be events with P(A)=0.7 and P(B)=0.4, and P( A or B)= 0.8.
(a) Compute P ( A and B).
(b) Are A and B mutually exclusive? Explain
(c) Are A and B independent? Explain

Answer by ikleyn(52864) (Show Source): You can put this solution on YOUR website!
.


(a)  Use the general formula of the Probability theory

         P(A or B) = P(A) + P(B) - P(A and B).


     Substitute the given data into this formula.  You will get

         0.8 = 0.7 + 0.4 - P(A and B) 


     From this equation, find  

         P(A and B) = 0.7 + 0.4 - 0.8 = 0.3.


     It is the ANSWER  to question (a).



(b)  Since P(A and B) is not equal to 0 (zero, ZERO),  the events A and B  are NOR mutually exclusive.    ANSWER


     Notice :  for mutually exclusive events X and Y,  P(X and Y) = 0.




(c)  The events X and Y are called independent if  P(X and Y) = P(X)*P(Y).


     Let's check if it is true in our case.


     In (a), we found that  P(A and B) = 0.3.


     From the other side,  P(A)*P(B) = 0.7*0.4 = 0.28.


     Since 0.3 is not equal to 0.28, the events A and B are NOT independent.    ANSWER

Solved ---- all question are answered.

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