# Probability independent events 3

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)
=0.665+0.035-(0.665*0.035)
=0.676725


is it correct?

• 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" – Matti P. Apr 6 '18 at 7:35
• You mean for no.2 is it P(BUA')∩P(BUA)? – johnc Apr 6 '18 at 7:38
• 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? – johnc Apr 6 '18 at 7:44
• Actually in the second question, it doesn't matter whether $A$ occurs. The answer is just the probability of $B$. – Matti P. Apr 6 '18 at 7:46
• I think you have the symbols for intersection and union mixed up in your first comment. – Matti P. Apr 6 '18 at 7:47

• $$P(A\text{ and }\neg B)=P(A)P(\neg B) = P(A)(1-P(B)) = 0.05 \times (1-0.7) = 0.015$$
• $$\begin{split} P([\neg A\text{ and }B ]\text{ or }[A \text{ and } B]) &= P(\neg A\text{ and }B )+ P(A \text{ and } B) \\ &= (1-P(A))P(B) + P(A)P(B)\\ &= P(B) \\&= 0.7 \end{split}$$
• Do you mean if $A \Leftrightarrow \neg B$? – Matti P. Apr 6 '18 at 9:12