# Explaining a probability for a repeated drawing without replacement

Consider the following experience : a bag contains some balls, say 3 blue balls and 4 red balls. A ball is drawn from the bag, then a second one without replacement of the first.

It's a classic result that the probability of drawing a blue ball the second time is the same as if the first ball had been replaced in the bag : $${\rm P}(B_2) = {\rm P}(B_1){\rm P}(B_2|B_1) + {\rm P}(\overline{B_1}){\rm P}(B_2|\overline{B_1}) = \frac37\frac26 + \frac47\frac36 = \frac37$$ My question is : how do I explain to my students why the probability doesn't change ? I have a computation to answer the question, but no "common probability sense" answer.

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• As long as the first ball is not identified, the second ball probability doesn't matter whether or not the first ball is put back. 7 balls (1 outside or not) 3 blue. Commented Mar 14, 2020 at 17:29
• Generalization. $n$ balls, $m$ blue. Split $n$ balls into 2 or more piles (without looking). Pick 1 ball from any pile, the blue probability is $\frac{m}{n}$. Commented Mar 14, 2020 at 17:36
• Each of the seven balls is equally likely to be the 2nd ball drawn, in both situations.
– Ned
Commented Mar 15, 2020 at 17:02

Here is a way I think many students would have an easy time understanding.

Think if we were to apply the numbers $$1$$ through $$7$$ to the balls at random. Perhaps we draw them on the balls with a marker. We could put the numbers on in whichever order we wanted, so long as the numbers are applied randomly. If we were to apply #$$2$$ first, we obviously have a $$3/7$$ chance that #$$2$$ goes to a blue ball.

This applies to any #$$k$$, the probability that the $$k$$th ball will be blue is still $$3/7$$.

Selecting the balls without replacement defines a sequence of blue and red balls. Thus, the probability that the second ball we select is blue is equal to the probability that the second ball in a random sequence of three blue and four red balls is blue. Since the three blue balls are equally likely to be in any of the seven positions in the sequence, the probability that the second ball is blue is $$3/7$$.

Well, the two events are equally likely and in general, equally likely events do not have a "common probability sense".

$$P(R_1)P(\frac{B_2}{R_1})+P(R_1)P(\frac{B_2}{R_1})=P(R_1 \cap B_2) + P(B_1 \cap B_2)$$

$$=P(B_2 \cap R_1) + P(B_2 \cap B_1)=P(B_2)P(\frac{B_2}{B_1})+P(B_2)P(\frac{B_1}{B_2})$$

$$=P(B_2)\left ( P(\frac{R_1}{B_2})+P(\frac{B_1}{B_2})\right)= P(B_2)$$

It is just that the probability of getting a second ball blue is same as the probability of drawing the first ball blue and that the two events are equally likely.