SOLUTION: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls

Question 1177712: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls do you expect to be left in the box? (Hint: find the expected value of the properly defined random variable)
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Let's solve this problem by considering the probability of each red ball remaining in the box.
**Understanding the Problem**
We have 3 red balls and 4 black balls. We draw balls without replacement until all 4 black balls are drawn. We want to find the expected number of red balls remaining.
**Reasoning**
A red ball will remain in the box if it is drawn *after* all the black balls have been drawn. This means that a red ball remains if the last black ball is drawn *before* that red ball is drawn.
Consider the positions of the balls in the sequence of draws. We have a total of 7 balls (3 red, 4 black).
Let's look at a specific red ball. It will remain in the box if the last black ball is drawn before it.
Consider any specific red ball and the last black ball. There are two possibilities:
1. The last black ball is drawn before the red ball.
2. The red ball is drawn before the last black ball.
Since the draws are random, these two possibilities are equally likely. Therefore, the probability that a specific red ball remains in the box is 1/2.
**Expected Value**
Let X be the number of red balls remaining in the box. We have 3 red balls.
* E(X) = (probability that red ball 1 remains) + (probability that red ball 2 remains) + (probability that red ball 3 remains)
* E(X) = 1/2 + 1/2 + 1/2 = 3/2 = 1.5
**Therefore, the expected number of red balls remaining in the box is 1.5.**

Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in.
Stop as soon as all the black balls are drawn out of the box.
How many red balls do you expect to be left in the box?
(Hint: find the expected value of the properly defined random variable)
~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @CPhill is CONCEPTUALLY WRONG, since it uses a wrong logic.
        Below I placed my CORRECT solution.

Let X be the number of remaining red balls in the box after stopping in this experiment. I will consider all possible outcomes for all cases X = 0, 1, 2, 3 separately. Case X = 3. It means that in the experiment, we draw only black balls: all 4 black balls, one after another. The probability for this case is P = = 0.028571429. Case X = 2. It means that in the experiment, 4 black balls were drawn and 1 red ball; the 4th black ball was the last, which was drawn. Let's use letter B to show black ball drawn and letter R to show red ball drawn. Then the following four outcomes are possible RBBBB, BRBBB, BBRBB, BBBRB with B as the last letter. (5-1 = 4 possible placements for R among 5 possible positions). The probabilities for these outcomes are RBBBB: P = = 0.028571429 BRBBB: P = = 0.028571429 BBRBB: P = = 0.028571429 BBBRB: P = = 0.028571429 Interesting fact is that we got 4 equal probabilities. We did not assume it for advance. What we got, is entirely experimental fact. Case X = 1. It means that in the experiment, 4 black balls were drawn and 2 red balls; the 4th black ball was the last, which was drawn. Then the following C(5,2) = 10 outcomes are possible RRBBBB, RBRBBB, RBBRBB, RBBBRB, | BRRBBB, BRBRBB, BRBBRB, BBRRBB, BBRBRB, | BBBRRB with B as the last letter. ( here C(5,2) = 10 means 10 possible placements for two B's in 5 possible positions). The probabilities for these outcomes are RRBBBB: P = = 0.028571429 RBRBBB: P = = 0.028571429 . . . I will not continue with the remaining 10-2 = 8 cases and will accept without a proof that in all 8 remaining cases the probability has the same value of 0.028571429. Later I will check this assumption. Case X = 0. It means that in the experiment, 4 black balls were drawn and 3 red balls; the 4th black ball was the last, which was drawn. Then the following C(6,3) = 20 outcomes are possible, that are 20 7-letter words with 4B, 3R and B in the last position I will tell without the proof that the probability is 0.028571429 in all/each of these cases. CHECKING my conception. P(overall) = 0.028571429 + 4*0.028571429 + 10*0.028571429 + 20*0.028571429 = 1.0000000. It means that everything is CORRECT with my calculations and my suggestions, so far. Now the expectation for the number of the red balls remained in the box is E = 3*0.028571429 + 2*(4*0.028571429) + 1*(10*0.028571429) + 0*(20*0.028571429) = 0.6. ANSWER. The expectation for the number of red ball remained in the box is 0.6.

Solved correctly.

--------------------

The logic in the solution by @CPhill is thoroughly false and thoroughly rotten.

There is a joke in the Probability theory that every event has the probability 1/2 = 0.5: it ether happens or not.

The logic in the solution by @CPhill literally and precisely follows to this joking formula,
with a completely serious expression on his face and without any smile, which tells about extreme degree of stupidity.

/////////////////////////////////////////////

                Regarding the post by @CPhill . . .

Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.

The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.

                It has no feeling of shame - it is shameless.

This time, again, it made an error.

Although the @CPhill' solution are copy-paste Google AI solutions, there is one essential difference.

Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.

All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.

Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.

And the last my comment.

When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.

Without it, their reliability is ZERO and their creadability is ZERO, too.

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