# probability of drawing marbles based on coin flip

An urn contains 4 red and 3 white marbles. A random marble $M_1$ is drawn and a fair coin is flipped. If the flip is heads then $M_1$ is put back into the urn. On the other hand, if the flip is tails, the marble $M_1$ is not put back into the urn. Now another random marble $M_2$ is drawn from the urn.

(i) What is $\mathrm{Pr}(M_2 = \text{red})$?
(ii) What is $\mathrm{Pr}(M_1 = \text{red}\mid M_2 = \text{red})$?
(iii) What is $\mathrm{Pr}(\text{flip is heads}\mid M_2 = \text{white})$?

Solutions:  (i) $\dfrac{4}{7}$;   (ii) $\dfrac{15}{28}$;   (iii) $\dfrac{1}{2}$

(iii)'s answer is a half because the the marble color is independent of the coin flip, correct?

Why is (i)'s answer $\dfrac{4}{7}$, why not $\dfrac{3}{6}$ or $\dfrac{4}{6}$ if the coin flip outcome is tails?

As for (ii), I'm completely lost.

Can someone explain please?

(i) The probability that $M_2$ is red depends on what happened on the first draw. Consider three events: the coin comes up heads ($H$); $M_1$ is red and the coin comes up tails ($R$); $M_1$ is white and the coin comes up tails ($W$). Then $\Pr(H)=\frac12$, $\Pr(R)=\frac12\cdot\frac47=\frac27$, and $\Pr(W)=\frac12\cdot\frac37=\frac3{14}$. The probability that $M_2$ is red is $\frac47$ if $H$ occurred, $\frac36=\frac12$ if $R$ occurred, and $\frac46=\frac23$ if $W$ occurred, so $$\Pr(M_2\text{ is red})=\frac12\cdot\frac47+\frac27\cdot\frac12+\frac3{14}\cdot\frac23=\frac27+\frac17+\frac17=\frac47\;.$$

(ii) You can use Bayes’ theorem, but you can also think it through without using any formula, and it’s not a bad idea to do so at least once. I already did much of the necessary calculation in answering (i). Informally, we know from (i) that on average $M_2$ is red $4/7$ of the time; we want to know in what fraction of that $4/7$ $M_1$ was also red. In (i) we didn’t care what $M_1$ was when the coin came up heads, because we put it back, and its color didn’t affect the probability that $M_2$ was red. Now we do care, so we split the event $H$ into $RH$ and $WH$: $RH$ is the event that $M_1$ is red and the coin comes up heads, and $WH$ is the event that $M_1$ is white and the coin comes up heads. Clearly $\Pr(RH)=\frac12\cdot\frac47=\frac27$, and $\Pr(WH)=\frac12\cdot\frac37=\frac3{14}$. The probability that $M_2$ is red if $RH$ occurred is $\frac27\cdot\frac47=\frac8{49}$, and the probability that $M_2$ is red if $WH$ occurred is $\frac3{14}\cdot\frac47=\frac6{49}$. We now have the following probabilities:

$$\begin{array}{cl|l} &\text{Event}&\text{Probability}\\ \hline *&M_1\text{ is red and }M_2\text{ is red and coin is heads}&\frac8{49}\\ &M_1\text{ is white and }M_2\text{ is red and coin is heads}&\frac6{49}\\ *&M_1\text{ is red and }M_2\text{ is red and coin is tails}&\frac17\\ &M_1\text{ is white and }M_2\text{ is red and coin is tails}&\frac27\\ \hline &\text{total probability that }M_2\text{ is red}&\frac47 \end{array}$$

The starred lines show the cases in which $M_1$ is red. As you can see, they contribute a total of $\frac8{49}+\frac17$ to the total probability of $\frac47$ that $M_2$ is red. This means that

$$\Pr(M_1\text{ is red}\mid M_2\text{ is red})=\frac{\frac8{49}+\frac17}{\frac47}=\frac{\frac{15}{49}}{\frac47}=\frac{15}{28}\;.$$

(iii) Once $M_1$ has been drawn and the coin flipped, the probability that $M_2$ is white depends on the outcome of the coin flip, since it depends on whether $M_1$ was replaced or not. Thus, you have to make a calculation similar to the one in (ii). I’ll just outline it:

$$\begin{array}{cl|l} &\text{Event}&\text{Probability}\\ \hline *&M_1\text{ is red and }M_2\text{ is white and coin is heads}&\frac47\cdot\frac37\cdot\frac12=\frac6{49}\\ *&M_1\text{ is white and }M_2\text{ is white and coin is heads}&\frac37\cdot\frac37\cdot\frac12=\frac9{98}\\ &M_1\text{ is red and }M_2\text{ is white and coin is tails}&\frac47\cdot\frac12\cdot\frac12=\frac17\\ &M_1\text{ is white and }M_2\text{ is white and coin is tails}&\frac37\cdot\frac13\cdot\frac12=\frac1{14}\\ \hline &\text{total probability that }M_2\text{ is red}&\frac37 \end{array}$$

Now the starred cases contribute $\frac6{49}+\frac9{98}=\frac3{14}$ to the total probability of $\frac37$ that the coin comes of heads, so

$$\Pr(\text{coin is heads}\mid M_2\text{ is white})=\frac{3/14}{3/7}=\frac12\;.$$

(i) Every marble is just as likely to be picked for $M_2$ as any other marble. That is why the probability is $\frac{4}{7}$. If you want to see this in more detail, lets look at all the possible ways of getting $M_2$ = red. Flip heads: white then red, red then red, flip tails: White then red, red then red. Total probability:

$$\frac{1}{2} \frac{3}{7} \frac{4}{7} + \frac{1}{2} \frac{4}{7} \frac{4}{7} + \frac{1}{2} \frac{3}{7} \frac{4}{6} + \frac{1}{2} \frac{4}{7} \frac{3}{6} = \frac{4}{7}$$

(ii) Using $P(A|B) = P(A \cap B) / P(B)$

$$P(M_1 = \text{red} \cap M_2 = \text{red}) = \frac{1}{2} \frac{4}{7} \frac{3}{6} + \frac{1}{2} \frac{4}{7} \frac{4}{7} = \frac{15}{49}$$

Giving $\frac{15}{49}/\frac{4}{7} = \frac{15}{28}$ as the answer.

(iii) $$P(\text{flip = heads} \cap M_2 = \text{white}) = \frac{1}{2} \left(\frac{4}{7} \frac{3}{7} + \frac{3}{7} \frac{3}{7}\right) = \frac{3}{14}$$

$P(M_2 = \text{white}) = \frac{3}{7}$ by similar argument to (i) and so using the conditional probability rule, the final answer is $\frac{3}{14} / \frac{3}{7} = \frac{1}{2}$

(i)

Given there are 4 red and 3 white marbles, there is a $\frac 4 7$ probability that $M_1 =$ red is drawn and a $\frac37$ probability that $M_1 =$ white. Now here is the tricky bit. A fair coin is flipped.

• If the coin flip is heads, then the urn will contain 4 red and 3 white for the 2nd draw.
• If the coin flip is tails, then there is a $\frac47$
• probability that the urn will contain 3 red and 3 white; and a $\frac37$ probability that the urn will contain 4 red and 2 white.

Both these events have a $50/50$ probability of occurring. So taking into account all this

$$P(M_2 \text{is red}) = 0.5\begin{pmatrix}\frac47\end{pmatrix} + 0.5\begin{bmatrix}\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac36\end{pmatrix} +\begin{pmatrix}\frac37\end{pmatrix}\begin{pmatrix}\frac46\end{pmatrix}\end{bmatrix} = \frac47$$

(ii)

$$P(M_1 \text{is red} \cap M_2 \text{is red})\\= 0.5\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac47\end{pmatrix} + \begin{bmatrix} 0.5\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac36\end{pmatrix}\end{bmatrix}=\frac{15}{49}$$

Taking the conditional probability formula,

\begin{align*} P(M_1 \text{is red} | M_2 \text{is red}) &= \frac{P(M_1 \text{is red} \cap M_2 \text{is red})}{P( M_2 \text{is red})}\\ &= \frac{\frac{15}{49}}{\frac47}\\ &=\frac{15}{28} \end{align*}