# 4.4 Random Variables, Density Functions

## A Random Number

Here is a process to construct a random number:

1. 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.

1. What is $$\Pr[N=0]$$?

Exercise 1

The only way to get 0 is if you first flip tails and then flip 0 heads. Therefore,
$$\Pr[N=0] = \Pr[F=0 \text{ AND } 0 \text{ heads}] = \Pr[F=0]\Pr[0 \text{ heads}] = \frac{2}{5}\cdot\frac{1}{8}=\frac{1}{20}$$.

2. What is $$\Pr[N=3]$$?

Exercise 2

The only way to get 3 is if you first flip heads and then roll a 3. Therefore,
$$\Pr[N=3] = \Pr[F=1 \text{ AND roll } 3] = \Pr[F=1] \Pr[\text{roll } 3] = \frac{3}{5}\cdot \frac{1}{6} = \frac{1}{10}$$.

3. What is $$\Pr[N=6]$$?

Exercise 3

You can get 6 either by flipping heads and rolling a 6 or by flipping tails and then flipping 3 heads. Therefore,
$$\Pr[N=6] = \Pr[F=1 \text{ AND roll } 6] + \Pr[F=0 \text{ AND } 3 \text{ heads}] = \frac{3}{5}\cdot \frac{1}{6} + \frac{2}{5}\cdot \frac{1}{8} = \frac{3}{20}$$.

4. What is $$\Pr[N=7]$$?

Exercise 4

There is no way to end up with the number 7.

5. What is $$\Pr[N=6\;|\;F=0]$$?

Exercise 5

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 $$\frac{1}{8}$$.

6. What is $$\Pr[F=0\;|\;N=6]$$?

Exercise 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.

7. What is $$\Pr[N+F=5]$$?

Exercise 7

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,
$$\Pr[N+F=5] = \Pr[F=1 \text{ AND } N=4] = \Pr[F=1] \Pr[\text{roll } 4] = \frac{3}{5}\cdot \frac{1}{6} = \frac{1}{10}$$.

8. What is $$\Pr[N+F=6]$$?

Exercise 8

This can happen either because $$F=1$$ and $$N=5$$ or because $$F=0$$ and $$N=6$$.
$$\Pr[N+F=6] = \Pr[F=1 \text{ AND } N=5] + \Pr[F=0 \text{ AND } N=6] = \frac{3}{5}\cdot \frac{1}{6} + \frac{2}{5}\cdot \frac{1}{8} = \frac{3}{20}$$.