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Let me take this question:

 The probability of raining tomorrow is 0.2. 
 Also, tomorrow I will toss a fair coin.
 What is the probability that tomorrow it rains and I get a head in the coin toss?

(Let's take that these two events are independent.)

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  • $\begingroup$ If $A$ and $B$ are independent events, then $P(A\cap B)=P(A)P(B),$ because that's the definition of independent events. $\endgroup$ – bof Dec 10 '17 at 10:10
  • $\begingroup$ But, why? Before, defining that definition, they must have had some intuition, or some way to prove it. $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 10:13
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Assume that we have an experiment with $M$ equiprobable outcomes $x_k$, $m$ of them considered favorable, and there is a second experiment with $N$ equiprobable outcomes $y_l$, whereby $n$ of them are considered favorable. The probabilities of success in these two experiments then are ${m\over M}$ and ${n\over N}$, respectively.

Calling these two experiments independent means, by definition, that the $MN$ possible combined outcomes $(x_k,y_l)$ are considered equiprobable. Among these $MN$ combined outcomes there are $mn$ where both experiments turn out successfully. The probability that this happens is $${mn\over MN}={m\over M}\cdot{n\over N}\ .$$

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  • $\begingroup$ Yes, that's it!! Now I get it! $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 11:48
  • $\begingroup$ I have one more doubt. "Tossing a coin", can be divided into two equiprobable outcomes: head or tail. So in our case, getting a head is the only outcome considered favorable. Thus, giving it a probability(of success) of 1/2. But in the case of "will it rain tomorrow", how do you divide it into equiprobable outcomes? $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 12:04
  • $\begingroup$ Experience tells us that in two out of ten days with the present atmospheric conditions it will rain tomorrow. $\endgroup$ – Christian Blatter Dec 10 '17 at 15:35
  • $\begingroup$ Hmm.........Okay!!!! $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 16:16
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Half of the time that it rains, you'll get tails rather than heads (assuming independence and a fair coin, of course.) So it will only rain and come up heads half as often as it rains. So we multiply the probability of rain by the probability of heads (one half).

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  • $\begingroup$ What do you mean by 'time'? $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 10:04
  • $\begingroup$ Assuming you hypothetically repeated tomorrow over and over. (Or say you had a bunch of days with twenty percent chance of rain and you flipped a coin on each one.) $\endgroup$ – spaceisdarkgreen Dec 10 '17 at 10:12
  • $\begingroup$ Now, I am confused by the meaning of "probability of raining tomorrow is 20%". What does it mean? $\endgroup$ – L.Shane John Paul Newton Dec 10 '17 at 10:20
  • $\begingroup$ That is a philosophical question with no widely agreed-upon answer en.wikipedia.org/wiki/Probability_interpretations $\endgroup$ – spaceisdarkgreen Dec 10 '17 at 10:22
  • $\begingroup$ Start by looking at non-frequentist interpretations, such as Bayesian, Dempster-Shafer, et al. $\endgroup$ – E. Douglas Jensen Dec 11 '17 at 2:25

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