# How to apply probability for flipping a coin and tossing a die? [closed]

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?

$P(\text{rolling a number on a die}) = \frac{1}{6}$

## closed as off-topic by Ethan Bolker, Delta-u, B. Mehta, Mr Pie, José Carlos SantosMay 20 '18 at 6:53

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• You have asked dozens of questions on this site in the last three months. Many have answers, but you haven't accepted any answers. That's not polite. Moreover, many of your questions, like this one, show no work of your own. That's not the accepted way to ask here. – Ethan Bolker May 19 '18 at 21:11
• Sorry about that, I'll keep that in mind for the future – dg123 May 19 '18 at 21:48

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 $\frac 16$.

• So it's basically asking for at least a 3? – dg123 May 19 '18 at 21:49
• In order to get $3$ heads in some amount of coin tosses, you have to flip the coin at least three times, therefore rolling a $1$ or $2$ will not allow you enough flips – Rhys Hughes May 19 '18 at 21:53
• And does 4C3, 5C3 and 6C3 mean the number on the die choosing 3 heads? – dg123 May 19 '18 at 21:57
• $4C3$ is the number of ways you can get 3 heads from 4 coin tosses: Note $4C3=4$ and you can have $HHHT, HHTH, HTHH, THHH$ as possible ways. – Rhys Hughes May 19 '18 at 21:58
• So 5C3 and 6C3 mean the number of ways you can get 3 heads from 5 and 6 tosses? – dg123 May 19 '18 at 21:59