Answer:
(a) Sample Space
[tex]S = \{0,1,2,3\}[/tex]
(b) PMF
[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}[/tex]
(c) CDF
[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}[/tex]
Step-by-step explanation:
Solving (a): The sample space
From the question, we understand that at most 3 cars will be repaired.
This implies that, the number of cars will be 0, 1, 2 or 3
So, the sample space is:
[tex]S = \{0,1,2,3\}[/tex]
Solving (b): The PMF
From the question, we have:
[tex]P(2) = P(1)[/tex]
[tex]P(0) = P(3)[/tex]
[tex]P(1\ or\ 2) = 0.5 * P(0\ or\ 3)[/tex]
[tex]P(1\ or\ 2) = 0.5 * P(0\ or\ 3)[/tex] can be represented as:
[tex]P(1) + P(2) = 0.5[P(0) + P(3)][/tex]
Substitute [tex]P(2) = P(1)[/tex] and [tex]P(0) = P(3)[/tex]
[tex]P(1) + P(1) = 0.5[P(0) + P(0)][/tex]
[tex]2P(1) = 0.5[2P(0)][/tex]
[tex]2P(1) = P(0)[/tex]
[tex]P(0)= 2P(1)[/tex]
Also note that:
[tex]P(0) + P(1) + P(2) + P(3) = 1[/tex]
Substitute [tex]P(2) = P(1)[/tex] and [tex]P(0) = P(3)[/tex]
[tex]P(0) + P(1) + P(1) + P(0) = 1[/tex]
[tex]2P(1) + 2P(0) = 1[/tex]
Substitute [tex]P(0)= 2P(1)[/tex]
[tex]2P(1) + 2*2P(1) = 1[/tex]
[tex]2P(1) + 4P(1) = 1[/tex]
[tex]6P(1) = 1[/tex]
Solve for P(1)
[tex]P(1) = \frac{1}{6}[/tex]
To calculate others, we have:
[tex]P(2) = P(1)[/tex]
[tex]P(2) = P(1) = \frac{1}{6}[/tex]
[tex]P(0)= 2P(1)[/tex]
[tex]P(0) =2 * \frac{1}{6}[/tex][tex]P(0) =\frac{1}{3}[/tex]
[tex]P(3) = P(0) =\frac{1}{3}[/tex]
Hence, the PMF is:
[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}[/tex]
See attachment (1) for histogram
Solving (c): The CDF ; F(x)
This is calculated as:
[tex]F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)[/tex]
For x = 0;
We have:
[tex]P(X \le 0) = P(0)[/tex]
[tex]P(X \le 0) = 1/3[/tex]
For x = 1
[tex]P(X \le 1) = P(0) + P(1)[/tex]
[tex]P(X \le 1) = 1/3 + 1/6[/tex]
[tex]P(X \le 1) = 1/2[/tex]
For x = 2
[tex]P(X \le 2) = P(0) + P(1) + P(2)[/tex]
[tex]P(X \le 2) = 1/3 + 1/6 + 1/6[/tex]
[tex]P(X \le 2) = 2/3[/tex]
For x = 3
[tex]P(X \le 3) = P(0) + P(1) + P(2) + P(3)[/tex]
[tex]P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3[/tex]
[tex]P(X \le 3) = 1[/tex]
Hence, the CDF is:
[tex]\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}[/tex]
See attachment (2) for histogram
