Question 2.2.1
A gardener planted three types of bean seeds: 35 green beans, 10 yellow beans, and 5 magic beans. Mature green bean plants sell for $5, yellow bean plants sell for $7, and magic bean plants sell for $50.
Unfortunately the plants look the exact same when matured. The farmer randomly selects a plant to sell.
Define the random variable \(Z\) as the amount of money made off of the sale of a plant. Calculate the distribution of \(Z\) for any \(z \in {\mathbb{R}}\).
There are three possible amounts of money the farmer can make: $5, $7, or $30. The probabilities selling the plant associated with the price is 35/50, 10/50, and 5/50, respectively.
This means:
\[
\mathbb{P}(Z = z) = \begin{cases}
35/50 & z=5\\
10/50 & z = 7\\
5/50 & z = 30 \\
0 & \text{otherwise}
\end{cases}
\]
Question 2.2.2
Consider flipping three fair coins. Let \(X\) be the number of heads showing. Write out the distribution of \(X\) for any \(x \in {\mathbb{R}}\).
\(\Omega = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}\)
There is one event with no heads.
There are 3 events with exactly one head.
There are 3 events with exactly two heads
There is 1 event with exactly three heads.
Each of these events has a probability of 1/8. Thus:
\[
P(X = x) = \begin{cases}
1/8 & x = 0\\
3/8 & x = 1 \\
3/8 & x = 2\\
1/8 & x= 3\\
0 & \text{otherwise}
\end{cases}
\]
Question 2.2.3
Suppose we roll one fair six-sided die. Let \(X\) be the number showing and let \(Z = \max(4, X)\).
Compute \(P(Z = z)\) for any \(z \in {\mathbb{R}}\).
The probability of rolling any number is 1/6. If we roll a 1, 2, 3, or 4, then \(Z = \max(4, X) = 4\). The probability of rolling a 1, 2, 3, or 4 is 4/6 or 2/3.
Otherwise, if we roll a 5, \(Z = \max(4, 5) = 5\) and similarly \(Z = \max(4, 6) = 6\). Each of these events have a probability of 1/6.
Thus:
\[
P(Z = z) = \begin{cases}
2/3 & z = 4\\
1/6 & z = 5 \\
1/6 & z = 6\\
0 & \text{otherwise}
\end{cases}
\]