# Is this probability question wrong?

Suppose we have the following information:

1.There is a 60 percent chance that it will rain today.

2.There is a 50 percent chance that it will rain tomorrow.

3.There is a 30 percent chance that it does not rain either day.

Find the following probabilities: The probability that it will rain today or tomorrow. The probability that it will rain today and tomorrow. The probability that it will rain today but not tomorrow. The probability that it either will rain today or tomorrow, but not both.

Shouldn't the probability that it does not rain either day be (1-0.60)*(1-0.50)=0.20 ? Is there an inconsistency here?

• You are assume the rain is independent. Maybe if it rains today the probability it will rain the next day goes up (or down). Here's another example. The probability I win \$200 in the office pool is$15\%$. The probability I but buy a steak dinner tomorrow is$20\%$. The probability I do both is$18\%$. Why isn't$(1-.15)(1-.20)=17\%\$? Because if I win I will figure I can afford it and will be more likely to by steak than if I don't. Feb 2, 2020 at 8:21

There is no inconsistency here. Your calculation is making the additional hypothesis that the events $$A=\{\textrm{it will rain today}\}$$ and $$B=\{\textrm{it will rain tomorrow}\}$$ are independent. But as any weather man will tell you, these events are far from independent - they can have correlation.
Thus in symbols, $$P(A)=.6,\ P(B)=.5,\ P(A^c\cap B^c)=.3,$$ and the question is asking you to compute these probabilities: $$P(A\cup B),\ P(A\cap B),\ P(A\setminus B),\ P\bigl((A\setminus B)\cup (B\setminus A)\bigr).$$
If $$A$$ and $$B$$ were independent, then as you point out, it would force $$P(A^c\cap B^c)=P(A^c)P(B^c)=(1-.6)(1-.5)=.2,$$ but we see that they are not independent, but rather they are positively correlated since $$.3=P(A^c\cap B^c)>P(A^c)P(B^c)=.2$$