## more Probability problems

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

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

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

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?
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 Recognitions: Homework Help Science Advisor They are; except can you explain P(1st but not 2nd) = 0.7?
 yeah thats incorrect. It should be $$P(A \cap B') = P(A) - P(A \cap B) = 0.4-0.3 = 0.1$$

## more Probability problems

 Quote by tronter yeah thats 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

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