# A coin is flipped when dice hits 6 (conditional probability)

Prof gave us homework on conditional probability that is due on the day of the lecture on conditional probability. Yeah, this has been a bad week and I've no idea what I'm doing.

Q: 3 dice are rolled, then, a coin is flipped as many times as the number 6 is obtained.

a) find the probability of getting less than 2 heads.

b) knowing this experiment results in less than 2 heads, what is the conditional probability that exactly 2 sixes were obtained?

I don't even know where to start...

Attempt:

a)

The only way I figure is:

Rolling one six, probability of head < 2 = $\displaystyle \frac{3}{6^3} \cdot \frac{2}{2}$

Rolling two sixes: $\displaystyle \frac{3}{6^3} \cdot \frac{2}{2^2}$

Rolling three sixes: $\displaystyle \frac{1}{6^3} \cdot \frac{4}{2^3}$

b)

$\displaystyle \frac{\frac{1}{6^3} \cdot \frac{4}{2^3}}{\frac{1}{6^3}}$

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a) The computation takes a while. It may be useful to draw a tree in order not to lose track of the possibilities. Initially, we toss $3$ dice. We get $0$, $1$, $2$, or $3$ $6$'s. Then, depending on the outcome, we toss a certain number of coins. So from the "start" position, there are $4$ branches, corresponding to the number of $6$'s obtained.

If $3$ dice are rolled, then the probability of $k$ $6$'s is equal to $$\binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3-k}.$$ In particular, the probability of $0$ $6$'s is $\frac{125}{216}$, the probability of exactly $1$ $6$ is $\frac{75}{216}$, the probability of exactly $2$ $6$'s is $\frac{15}{216}$, and the probability of $3$ $6$'s is $\frac{1}{216}$.

If we have $0$ $6$'s, we toss $0$ coins, so for sure (probability $1$) we get fewer than $2$ heads. If we get $1$ $6$, then we toss $1$ coin, and again for sure we get fewer than $2$ heads. If we get $2$ $6$'s, we toss $2$ coins, and the probability of fewer than $2$ heads is easily computed to be $3/4$. If we get $3$ $6$'s, we toss $3$ coins. By symmetry the probability of fewer than $2$ heads is then $1/2$. It follows that the overall probability of fewer than $2$ heads is $$\frac{125}{216}\cdot 1 + \frac{75}{216}\cdot 1 +\frac{15}{216}\cdot \frac{3}{4}+\frac{1}{216}\cdot\frac{1}{2}.$$ This may simplify to $\dfrac{847}{864}$. But check the arithmetic!

b) Let $E$ be the event that exactly $2$ sixes were obtained, and let $L$ be the event we got fewer than $2$ heads. We want $P(E|L)$.

There are various formulas that we could now use. We go for the simplest one, though using it is a bit less mechanical than the full Bayes formula. Note that $$P(E|L)P(L)=P(E \cap L).$$ This formula comes straight from the the definition of conditional probability. From the above formula, we see that we will be essentially finished once we find $P(E\cap L)$ and $P(L)$.

The probability $P(L)$ that we get fewer than $2$ heads has been computed in part (a). So we only need to find $P(E \cap L)$. In the solution of part (a), we already saw that $$P(E)=\binom{3}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^1=\frac{15}{216}=\frac{5}{72}.$$ Given that we got exactly $2$ $6$'s, we got to flip the coin twice, and thus the probability of fewer than $2$ heads is $3/4$, as shown by @Yuval Filmus. It follows that $$P(E\cap L)=P(L|E)P(E)=\left(\frac{3}{4}\right)\left(\frac{5}{72}\right)=\frac{5}{96}.$$ Now put the pieces together. I think that $P(E|L)=\dfrac{45}{847}$. Do check the arithmetic!

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Thanks. I didn't realize to factor in the $5^{k-1}$ dice combinations that are not sixes. Also, I see I did $P(L|E)$ instead of $P(E|L)$. – iDontKnowBetter Oct 5 '11 at 11:33
We have to be very careful in checking whether we are looking for the right conditional probability. They all sound alike. – André Nicolas Oct 5 '11 at 15:06

Your thinking is correct, though you're not calculating the probabilities right. As an example, if you roll two sixes, then you're tossing two coins; the probability of rolling two heads is $1/2 \cdot 1/2 = 1/4$, so the probability of rolling less than two heads is $3/4$.

As for the second part, look up Bayes' rule.

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the last one is an attempts at implementing it. Though we've not seen it in class. $P(A|B)$ where $A = \frac{4}{2^3}$ and $B= \frac{1}{6^3}$. And isn't their union simply $\frac{4}{6^3 2^3}$? – iDontKnowBetter Oct 4 '11 at 23:49