The discrete random variable \(X\) takes values 1, 2, 3, 4 and 5, and its probability distribution is defined as follows.
$$\mathrm{P}(X = x) = \begin{cases}
a & x = 1, \\
\frac{1}{2}\mathrm{P}(X = x - 1) & x = 2, 3, 4, 5, \\
0 & \text{otherwise,}
\end{cases}$$
where \(a\) is a constant.
- Show that \(a = \frac{16}{31}\). [2]
The discrete probability distribution for \(X\) is given in the table.
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| P\((X = x)\) | \(\frac{16}{31}\) | \(\frac{8}{31}\) | \(\frac{4}{31}\) | \(\frac{2}{31}\) | \(\frac{1}{31}\) |
- Find the probability that \(X\) is odd. [1]
Two independent values of \(X\) are chosen, and their sum \(S\) is found.
- Find the probability that \(S\) is odd. [2]
- Find the probability that \(S\) is greater than 8, given that \(S\) is odd. [3]
Sheila sometimes needs several attempts to start her car in the morning. She models the number of attempts she needs by the discrete random variable \(Y\) defined as follows.
$$\mathrm{P}(Y = y + 1) = \frac{1}{2}\mathrm{P}(Y = y) \quad \text{for all positive integers } y.$$
- Find P\((Y = 1)\). [2]
- Give a reason why one of the variables, \(X\) or \(Y\), might be more appropriate as a model for the number of attempts that Sheila needs to start her car. [1]