# Drawing balls from an urn

I'm preparing for my midterm and I'm having trouble connecting all the dots in the following problem:

From an urn containing 2 yellow balls, 3 blue balls and 22 red balls, one ball is drawn, found to be red and discarded. Then 5 balls are drawn and thrown away, their colors un-noticed. Given this information, what is the probability that if you now draw four balls from the urn, they are all red?

From this logic I have deduced that we have 6 cases after drawing the 5 balls: 0 red balls drawn, 1 red ball drawn, 2 red balls drawn, 3 red balls drawn, 4 and then 5.

I worked by summing the probabilities of all 6 cases and the answer was way off the actual solution (it should be 0.400).

(21P4 + 20P4 + 19P4 + 18P4 + 17P4 + 16P4)/(21P4))


• There are a lot of red herrings here. First of all, the first draw is nearly irrelevant. Just assume you started with $21$ red. Then since we have no information about the $5$ deleted balls, they change nothing. Each ball is equally likely to be amongst the $4$ chosen. Can you finish from here? – lulu Jun 20 '19 at 18:00
• As for your attempt, you can indeed break down by cases based on the number of red balls drawn in the second step, and then the probabilities within each case will combine... but they will do so after conditioning the value based on the probability of entering their respective cases... (Though doing so is incredibly tedious compared to the more straightforward solution alluded to by lulu) – JMoravitz Jun 20 '19 at 18:05
• Note: I can't really follow your calculation. You distinguish a number of cases, but you appear to simply assume that they are equally probable, which does not appear to be true. But even then, I don't understand your denominator. As it stands, your expression is obviously $>1$ which makes no sense. – lulu Jun 20 '19 at 18:05
• But what's the probability of it being red then if we don't know how many red balls have been chosen? – user683641 Jun 20 '19 at 18:07
• @O.Sinno It doesn't matter whether or not we know how many red balls were drawn in the second step and we could pretend that we had never pulled them out in the first place and the value obtained will be the same. This is effectively for the same reason that the second card drawn from a deck being a queen is still $1/13$ – JMoravitz Jun 20 '19 at 18:09

After the first ball is drawn, seen to be red, and discarded, there will be $$21$$ red balls remaining and $$5$$ non-red balls remaining (what the exact colors of each non-red ball is doesn't matter).

Next, since we draw some balls, and then discard them without looking at them, this is effectively the same as though we never drew them at all in the first place as it comes to calculating probabilities of the color of the next few draws. As such, it is much easier to completely ignore that this second step occurred at all.

Continuing then, we effectively have $$26$$ balls remaining, $$5$$ of which are non-red and the remaining $$21$$ of which are red. We ask what the probability that when selecting four of these we get only reds. This follows the hypergeometric distribution. You can use permutations if you like, but I find it easier to use binomial coefficients. The answer being:

$$\frac{\binom{21}{4}}{\binom{26}{4}}$$

Your approach wasn't a bad one, but you forgot something crucial in your calculations, and that is to have "conditioned" each term based on the probability of entering that specific case.

In more math heavy description, you have $$Pr(A) = Pr(A\mid B_1)Pr(B_1)+Pr(A\mid B_2)Pr(B_2)+\dots+Pr(A\mid B_n)Pr(B_n)$$ where $$B_1,B_2,\dots,B_n$$ form a partition of the sample space. You calculated each of $$Pr(A\mid B_1),Pr(A\mid B_2),\dots$$ and used these in your attempt but completely ignored each of $$Pr(B_1),Pr(B_2),\dots$$.

Correcting your approach then, we actually have something more like:

$$\frac{\binom{21}{0}\binom{5}{5}}{\binom{26}{5}}\times\frac{\binom{21}{4}}{\binom{21}{4}} + \frac{\binom{21}{1}\binom{5}{4}}{\binom{26}{5}}\times\frac{\binom{20}{4}}{\binom{21}{4}} + \frac{\binom{21}{2}\binom{5}{3}}{\binom{26}{5}}\times\frac{\binom{19}{4}}{\binom{21}{4}} + \dots + \frac{\binom{21}{5}\binom{5}{0}}{\binom{26}{5}}\times\frac{\binom{16}{4}}{\binom{21}{4}}$$

If you go through the effort of calculating each answer, you will find that they are the same. The second approach however, you will find much more tedious and so is unadvised.

• Thank you I finally get it! I also referred to the link you provided earlier and noticed that the two exercises need the same approach. – user683641 Jun 20 '19 at 18:25
• So the problem simplifies to $\frac{21}{26}*\frac{20}{25}*\frac{19}{24}*\frac{18}{23} = 0.400$ – RayDansh Jun 20 '19 at 18:27
• @JMoravitz: Nice explanation (+1), but I'll make one observation. The probability of drawing $4$ red balls after discarding $d\;$balls is independent of $d$, provided $0\le d\le 22$. – quasi Jun 20 '19 at 19:51
• @quasi, yes, there is a tacit assumption which I didn't write. "this is effectively the same as though we never drew them at all in the first place as it comes to calculating probabilities of the color of the next few draws (assuming there are enough balls left to draw)." I didn't think it needed to be said since it should be obvious that the probability that if you draw $100$ balls out of a bag with only one ball in it that they will all be red occurs with probability zero. (That, or you are a stage magician) – JMoravitz Jun 20 '19 at 19:54