Probability of drawing a yellow ball after 1 or 2 have been drawn.

Question

The question is as follows:

An opaque urn contains ten balls, five red, four yellow, and one green. Three balls are drawn from the urn in sequence but at random, without replacement.

What is the conditional probability of drawing a yellow ball given that a yellow ball has been drawn? (Note: “a yellow ball has been drawn” means at least one yellow ball has turned up, in the first draw, the second, or both.)

I have been trying to solve this problem using Bayes' Theorem the following way:

$ P(drawing\:yellow\:last\: |\: 1\: or\: 2\: drawn) = \frac{P(1\: or\: 2\: drawn\: | \:drawing \:yellow \:last)\cdot P(drawing \:yellow \:last)}{P(1 \:or \:2 \:drawn)} $

Where $P(1 \:or \:2\: drawn\: | \:yellow\: drawn \:last)$ will be equal to $P(1\: or\: 2 \:drawn)$ because it is not a conditional probability (this may be where I'm going wrong).

$$ P(1 \:or \:2 \:drawn) = 2(\frac{4}{10}\cdot\frac{6}{9} ) + (\frac{4}{10}\cdot\frac{3}{9} ) = .666666 $$

$$ P(drawing \:yellow \:last) = 2\cdot \frac{4}{10}\cdot\frac{6}{9} \cdot \frac{3}{8} + \frac{4}{10}\cdot\frac{3}{9}\cdot \frac{2}{8} =.23333 $$

Resulting in .233333 which is not the correct answer of precisely 1/2 in the solutions given to me by my TA.

This problem is driving me insane! If you could steer me in the right direction, I'd be super grateful. Thanks!

EDIT: Here is the TA solution:

Answer 1

Let $T$ be the event yellow on third, and let $A$ be the event at least one yellow in the first two trials. We want $\Pr(T|A)$. By the definition of conditional probability, we have $$\Pr(T|A)=\frac{\Pr(A\cap T)}{\Pr(A)}.\tag{1}$$

We compute the two probabilities on the right-hand side of (1).

The event $A$ can happen in $3$ ways, which we can call YN, NY, and YY. The sum of the probabilities of these is $$\frac{4}{10}\cdot \frac{6}{9}+\frac{6}{10}\cdot\frac{4}{9}+\frac{4}{10}\cdot\frac{3}{9}.$$ The event $A\cap T$ can also happen in $3$ ways, which we can call YNY, NYY, and YYY. The probability is therefore $$\frac{4}{10}\cdot \frac{6}{9}\cdot\frac{3}{8}+\frac{6}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}+\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8}.$$ Now we use (1) and in principle are finished, though you are probably expected to simplify. It does simplify, a lot.

Answer 2

The first two lines of your proof are wrong and everything goes down hill from there.

Firstly, you've applied Bayes' wrong. It goes: $\mathsf P(A\mid B) = \dfrac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B)}$
[which you've edited to correct; ignore]

Secondly, those probabilities are not equal, because one is conditional on the third draw being yellow.

---

Let $Y_1, Y_2, Y_3$ be the events of drawing a yellow ball on draw $1,2, 3$, respectively.

We wish to calculate $\mathsf P(Y_3\mid Y_1\cup Y_2)$.

$$\begin{align} \mathsf P(Y_3\mid Y_1\cup Y_2) & = \frac{\mathsf P\big((Y_1\cup Y_2)\mid Y_3\big)\cdot\mathsf P\big(Y_3\big)}{\mathsf P\big(Y_1\cup Y_2\big)} & \text{Bayes'} \\[1ex] & = \frac{\Big(1-\mathsf P\big((Y_1\cup Y_2)^c\mid Y_3\big)\Big)\cdot\mathsf P\big(Y_3\big)}{\Big(1-\mathsf P\big((Y_1\cup Y_2)^c\big)\Big)} \end{align} $$

Now, $\mathsf P((Y_1\cup Y_2)^c)$ is the probability of drawing two of the five not-yellow balls on both first two draws.

Also, $\mathsf P((Y_1\cup Y_2)^c\mid Y_3)$ is the probability of drawing two of the five not-yellow balls on both first two draws, when one of the yellow balls is reserved for the third draw.

Can you complete?

Answer 3

It doesn't seem like Bayes' Rule is a lot of help for this problem. Instead, just go by the definition of the conditional.

Let $A=$ "last ball yellow" and $B=$ "at least one of first two balls yellow"

$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

The total number of ways to draw the first two balls is: $$\binom{10}{2}\times 2!=90$$ (i.e., choose two of the ten balls and then choose which order to put them in)

Then, the number of ways to choose the first two balls so that we get at least one yellow is: $$\binom{4}{1}\binom{3}{1}+\binom{2}{1}\binom{4}{1}\binom{6}{1}=60$$ (i.e, either choose a yellow ball then another yellow ball or choose one of the two draws to be yellow, choose a yellow ball for it, then choose a non-yellow ball for the other draw)

So $$P(B)=\frac{60}{90}=\frac{2}{3}$$

Now, the total number of ways to draw the three balls is: $$\binom{10}{3}\times 3!=720$$ (i.e., choose three of the ten balls and then choose which order to put them in)

And the number of ways to choose the three balls where the last one is yellow and at least one of the first two is yellow is: $$\binom{4}{1}(\binom{3}{1}\binom{2}{1}+\binom{2}{1}\binom{3}{1}\binom{6}{1})=168$$ (i.e., choose a yellow ball for the third one. Then either choose another yellow ball then another yellow ball or choose one of the first two draws to be yellow, choose a yellow ball for it, then choose a non-yellow ball for the other of the first two draws)

So $$P(A\cap B)=\frac{168}{720}=\frac{7}{30}$$

Thus, $$P(A|B)=\frac{^7/_{30}}{^2/_3}=\frac{7}{20}=0.35$$