More Probability problems

[tronter]

$$P(\text{stop at first light signal}) = 0.4 = P(A)$$

$$P(\text{stop at second light signal}) = 0.5 = P(B)$$

$$P(\text{stop at least one of two signals}) = 0.6 = P(A \cup B)$$

Then $$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.3$$

$$P( \text{first signal but not second}) = P(A \cap B') = 1-0.3 = 0.7$$

$$P(\text{exactly one signal}) = P(A \cup B) - P(A \cap B) = P(A \Delta B) = 0.3$$


Are these correct?

 

[EnumaElish]

They are; except can you explain P(1st but not 2nd) = 0.7?

 

[tronter]

yeah that's incorrect. It should be $$P(A \cap B') = P(A) - P(A \cap B) = 0.4-0.3 = 0.1$$

 

[Jrb599]

yeah that's incorrect. It should be $$P(A \cap B') = P(A) - P(A \cap B) = 0.4-0.3 = 0.1$$

Shouldn't it be $$P(A \cap B') = P(A) * (1-P(B))$$

Assuming there independent