A Random Number
Here is a process to construct a random number:
 Flip a biased coin that comes up heads with probability \(\frac{3}{5}\).
 If you get heads, you roll a fair die and return the result.
 Otherwise, you flip a fair coin 3 times and return twice the number of heads.
Let \(N\) be the number that you return.
Let \(F\) be the indicator random variable for the first coin flip (1 if heads and 0 if tails).
Please answer in the form of a decimal with three significant figures.

What is \(\Pr[N=0]\)?
The only way to get 0 is if you first flip tails and then flip 0 heads. Therefore,\(\Pr[N=0] = \Pr[F=0 \text{ AND } 0 \text{ heads}] = \Pr[F=0]\Pr[0 \text{ heads}] = \frac{2}{5}\cdot\frac{1}{8}=\frac{1}{20}\). 
What is \(\Pr[N=3]\)?
The only way to get 3 is if you first flip heads and then roll a 3. Therefore,\(\Pr[N=3] = \Pr[F=1 \text{ AND roll } 3] = \Pr[F=1] \Pr[\text{roll } 3] = \frac{3}{5}\cdot \frac{1}{6} = \frac{1}{10}\). 
What is \(\Pr[N=6]\)?
You can get 6 either by flipping heads and rolling a 6 or by flipping tails and then flipping 3 heads. Therefore,\(\Pr[N=6] = \Pr[F=1 \text{ AND roll } 6] + \Pr[F=0 \text{ AND } 3 \text{ heads}] = \frac{3}{5}\cdot \frac{1}{6} + \frac{2}{5}\cdot \frac{1}{8} = \frac{3}{20}\). 
What is \(\Pr[N=7]\)?
There is no way to end up with the number 7. 
What is \(\Pr[N=6\;\;F=0]\)?
Given that \(F=0\), we know that the first coin flip was tails. So, we are in the case where the fair coin is flipped 3 times. In that case, we get 6 only by flipping 3 heads. The probability of this happening is \(\frac{1}{8}\). 
What is \(\Pr[F=0\;\;N=6]\)?
We use the definition of conditional probability to compute:\(\Pr[F=0\;\;N=6] = \dfrac{\Pr[F=0 \text{ AND } N=6]}{\Pr[N=6]} = \dfrac{\frac{2}{5}\cdot \frac{1}{8}}{\frac{3}{20}} = \frac{1}{3}\), (where, for the denominator, we used the answer to a previous question). Hence, if somebody else has run the process and ended up with 6, we know there is 1 in 3 chance that he had to flip the fair coin. 
What is \(\Pr[N+F=5]\)?
If \(F=0\), then we are flipping the fair coin and doubling the number of heads, so \(N\) is even, and therefore, \(N+F\) cannot be odd. Hence, the only way for \(N+F\) to be odd is if \(F=1\). Then \(N+F=5\) if and only if \(N=4\), which happens exactly when we roll a 4. Overall,\(\Pr[N+F=5] = \Pr[F=1 \text{ AND } N=4] = \Pr[F=1] \Pr[\text{roll } 4] = \frac{3}{5}\cdot \frac{1}{6} = \frac{1}{10}\). 
What is \(\Pr[N+F=6]\)?
This can happen either because \(F=1\) and \(N=5\) or because \(F=0\) and \(N=6\).\(\Pr[N+F=6] = \Pr[F=1 \text{ AND } N=5] + \Pr[F=0 \text{ AND } N=6] = \frac{3}{5}\cdot \frac{1}{6} + \frac{2}{5}\cdot \frac{1}{8} = \frac{3}{20}\).