Calculating Probability of Snow in Exactly One of Two Cities

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To calculate the probability of snowing in exactly one of two cities, with the first city having a snow probability of 0.4 and the second city 0.7, the approach involves determining P(A and 'not B') and P('not A' and B). The correct calculation is P(A) * P('not B') + P('not A') * P(B), which simplifies to (0.4)(0.3) + (0.6)(0.7). The initial subtraction of the overlapping probabilities was unnecessary since the events are mutually exclusive. The final probability of snow in exactly one city is 0.4896, and the test taker acknowledges receiving partial credit for their approach.
BrownianMan
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I had this question on a test:

The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities?

The way I approached it was to say that the probability that it snows in exactly one of the two cities is:

Let A denote probability of snow in the first city.
Let B denote probability of snow in the second city.

P(A and 'not B') or P('not A' and B) = (0.4)(0.3) + (0.6)(0.7) - (0.4)(0.3)(0.6)(0.7) = 0.4896

Is this correct?
 
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I don't think you need that part you are subtracting.
 
Yeah, I just realized that. The events P(A and 'not B') and P('not A' and B) are mutually exclusive.

The question was out of 6, so at least I'll get part marks.
 

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