3. 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.

1. Show that \(a = \frac { 16 } { 31 }\). The discrete probability distribution for \(X\) is given in the table.

   \(x\)	1	2	3	4	5
   \(\mathrm { P } ( X = x )\)	\(\frac { 16 } { 31 }\)	\(\frac { 8 } { 31 }\)	\(\frac { 4 } { 31 }\)	\(\frac { 2 } { 31 }\)	\(\frac { 1 } { 31 }\)

2. Find the probability that \(X\) is odd. Two independent values of \(X\) are chosen, and their sum \(S\) is found.
3. Find the probability that \(S\) is odd.
4. Find the probability that \(S\) is greater than 8 , given that \(S\) is odd. 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. $$\mathrm { P } ( Y = y + 1 ) = \frac { 1 } { 2 } \mathrm { P } ( Y = y ) \quad \text { for all positive integers } y .$$
5. Find \(\mathrm { P } ( Y = 1 )\).
6. 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.
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