How to apply probability for flipping a coin and tossing a die? I'm not sure how to calculate the probability for this question. Here is what I have so far. Can anyone please help me out?
Suppose we roll a fair six-sided die and then flip a number of fair coins equal to
the number showing on the die. (For example, if the die shows 4, then we flip 4 coins.
What is the probability that the number of heads equals 3?
$P(\text{rolling a number on a die}) = \frac{1}{6}$
 A: There are fewer cases where this works than not, so we'll work out probabilities of the ways it works.
Firstly, you have to roll at least a $3$ on the dice, so we'll ignore $1$ and $2$.
Suppose you roll a $3$. There is a $\frac 16$ chance of this, then you get three dice rolls and all have to be heads. For this probability, we take $\binom{3}{3}\cdot\big(\frac 12\big)^3=\frac 18$ (the chance of getting any outcome when rolling a die 3 times is the $(\frac 12)^3$, the $\binom{3}{3}$works out the number of outcomes which have three heads, in this case $1$). This means the chance of rolling a three and then getting three heads is $\frac 16\cdot\frac 18=\frac{1}{48}$
In case you've never seen it: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, where $n!=1\cdot 2\cdot 3\cdot ... \cdot n$. You can use $nCr$ on a calculator for this too.
Now if you roll a $4$, the chance of 3 heads is $\frac 16 \cdot \binom{4}{3}\cdot (\frac{1}{2})^4=\frac{2}{48}$
For $5$, we get: $\frac 16 \cdot \binom{5}{3}\cdot (\frac{1}{2})^5=\frac{5}{96}$
And for $6$ we get $\frac 16 \cdot \binom{6}{3}\cdot (\frac{1}{2})^6=\frac{5}{96}$
Summing these probabilities gives us $\frac 16$.
