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Suppose on any given day, $P(A) = 0.37$ and $P(B) = 0.21$, where $A$ and $B$ are events where:

$A$ = Event where the average daily temperature is below $10$ degrees

$B$ = Event where the daily rainfall is over $3$mm

Find the probability of at least one of these events occurring on any given day if there is a $63\%$ chance of the daily rainfall being over $3$mm, if the temperature is under $10$ degrees.

Here's what I've tried:

$$P(B \mid A) = 0.63 \\ P(B \cap A) = P(A) \cdot 0.63 \\ P(B \cap A) = 0.37 \cdot 0.63 = 0.2331$$

However, is this possible as the intersection of $B$ and $A$ exceeds the probability of $B$ itself.

Then I proceed with: \begin{align} P(A \cup B) &= P(A) + P(B) - P(B \cap A) \\ &= 0.37 + 0.21 -0.2331 \\ &=0.3469 \end{align}

Kindly let me know if my calculations and understandings are correct. Thank you!

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    $\begingroup$ You have correctly deduced that the information given in the problem is inconsistent. That implies that any attempted calculation based on the data in the problem will yield only nonsense. $\endgroup$ Aug 15, 2019 at 5:16

1 Answer 1

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In order to see where the inconsistency comes from, write the probability of both events happening in two different ways.

P(A)P(B|A) = P(B)P(A|B)

P(A|B)= P(A)P(B|A)/P(B)

P(A|B) = 0.37 * 0.63 / 0.21 = 1.11 > 1

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