Events $A$ and $B$ are independent. Suppose event $A$ occurs with probability 0.05 and event $B$ occurs with probability 0.70.

1.Compute the probability that $A$ occurs but $B$ does not occur.

2.Compute the probability that either $B$ occurs without $A$ occurring or $A$ and $B$ both occur.

now that the events are independent My answer for no.1= P(A)*P(B)=0.05*0.70=0.035

2.P(BUA')∩P(BUA)=(P(B)*P(A')) ∩ (P(B)*P(A))

            =(0.70*0.95) ∩ (0.70*0.05) 

is it correct?

  • $\begingroup$ In the first question, you forgot to take into account that $B$ does NOT occur. And the second question is fairly easy, it's just "B and not A or B and A" $\endgroup$ – Matti P. Apr 6 '18 at 7:35
  • $\begingroup$ You mean for no.2 is it P(BUA')∩P(BUA)? $\endgroup$ – johnc Apr 6 '18 at 7:38
  • $\begingroup$ P(BUA')∩P(BUA)=(P(B)*P(A')) ∩ (P(B)*P(A)) =(0.70*0.95) ∩ (0.70*0.05) =0.665+0.035-(0.665*0.035) =0.676725 Am I correct? $\endgroup$ – johnc Apr 6 '18 at 7:44
  • $\begingroup$ Actually in the second question, it doesn't matter whether $A$ occurs. The answer is just the probability of $B$. $\endgroup$ – Matti P. Apr 6 '18 at 7:46
  • $\begingroup$ I think you have the symbols for intersection and union mixed up in your first comment. $\endgroup$ – Matti P. Apr 6 '18 at 7:47
  • $$ P(A\text{ and }\neg B)=P(A)P(\neg B) = P(A)(1-P(B)) = 0.05 \times (1-0.7) = 0.015 $$
  • $$ \begin{split} P([\neg A\text{ and }B ]\text{ or }[A \text{ and } B]) &= P(\neg A\text{ and }B )+ P(A \text{ and } B) \\ &= (1-P(A))P(B) + P(A)P(B)\\ &= P(B) \\&= 0.7 \end{split} $$
  • $\begingroup$ How to solve this if the event A and B is mutually exclusive? $\endgroup$ – johnc Apr 6 '18 at 9:05
  • $\begingroup$ Do you mean if $A \Leftrightarrow \neg B$? $\endgroup$ – Matti P. Apr 6 '18 at 9:12
  • $\begingroup$ Events A and B are mutually exclusive. Suppose event A occurs with probability 0.74 and event B occurs with probability 0.22. 1.Compute the probability that A occurs but B does not occur. 2.Compute the probability that either A occurs without B occurring or A and B both occur. Answer for 1. P(A and ¬B)=P(A)+P(¬B)-P(A∩B)=0.74+0.78-(0.74*0.78)=0.9428 2.P([¬B and A] or [A and B])=P(¬B and A)+P(A and B)=0.78*0.74+0.74*0.22=0.74 Is it correct? $\endgroup$ – johnc Apr 6 '18 at 9:20

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