# Proving that $P(B|C \cap A^{c}) = P(B|C)$

I was solving the following question:

Most mornings, Victor checks the weather report before deciding whether to carry an umbrella. If the forecast is “rain,” the probability of actually having rain that day is 80%. On the other hand, if the forecast is “no rain,” the probability of it actually raining is equal to 10%.

During fall and winter, the forecast is “rain” 70% of the time and during summer and spring, it is 20%.

The probability of Victor missing the morning forecast is equal to 0.2 on any day in the year. If he misses the forecast, Victor will flip a fair coin to decide whether to carry an umbrella.

On any day of a given season, he sees the forecast, if it says “rain” he will always carry an umbrella, and if it says “no rain,” he will not carry an umbrella. Are the events “Victor is carrying an umbrella,” and “The forecast is no rain” independent? Does your answer depend on the season?

And my attempt to solve it is as following:

Let A be the event of the forecast is "rain".

Let B be the event of Victor carrying an umbrella.

Let C be the event of Victor misses the forecast.

From the statement we have that: $$P(A) = 0.2 \space or \space 0.7$$ $$P(B|A) = 0.8$$ $$P(B|A^{c}) = 0.1$$ $$p(C) = 0.2$$

So we must prove that B and A are independent events.

Since $$A^{c}$$ has non-zero probability we can prove independence by proving that:

$$P(B|A^{c}) = P(B)$$

From Total Probability Theorem, we can write: $$P(B) = P(C)P(B|C) + P(C^{c})P(B|C^{c})$$ Since $$P(B|C^{c}) = P(A)$$ (Because if he doesn't miss the forecast then he will carry an umbrella if the forecast is "rain"), then: $$\implies P(B) = (0.2)(0.5)+ (0.8)P(A)$$ $$\implies P(B) = 0.1 + (0.8)P(A)$$

Now Also from the Total Probability Theorem, we can write: $$P(B|A^{c}) = P(C)P(B|C \cap A^{c}) + P(C^{c})P(B|C^{c} \cap A^{c})$$

It's obvious that $$P(B|C \cap A^{c}) = P(B|C)$$, and from that It's easy to complete solving the problem.

My question is how we can prove that $$P(B|C \cap A^{c}) = P(B|C)$$.

Proving that: $$P(B| C \cap A^{c}) = P(B | C)$$ is tell us that the two events A and B are conditionally independent (that condition is C here).
Since B is determined by a random process If he misses the forecast, Victor will flip a fair coin to decide whether to carry an umbrella then we can say that B is independent from A since flipping a fair coin is not related by any way to A and have a fixed probability which equal to 0.5 .