Solutions

3.1

There are $n = N_{1}-N_{0}+1$ values. Let $Y=X-N_{0}+1$. Then from problem 2.24.b,

$$E[Y] = \frac{n+1}{2}$$

$$(Y) = \frac{n^{2}-1}{12}$$

So

$$E[X] = N_{0}-1+E[Y] = N_{0}-1+\frac{N_{1}-N_{0}+2}{2} = \frac{N_{1}+N_{0}}{2}$$

$$(X) = (Y) = \frac{(N_{1}-N_{0}+1)^{2}-1}{12}$$

3.3

Need:

• no cars in seconds 5, 6, 7--$(1-p)^{3}$
• one car in second 4 (otherwise could go earlier)--$p$
• at least one car in seconds 1, 2, 3-- $(1-(1-p)^{3})$

Probability:

$$(1-(1-p)^{3}) p (1-p)^{3}$$

3.7

$P(X \ge 2) = 0.99$ means

$$P(X=0)+P(X=1) = e^{-\lambda}+\lambda e^{-\lambda} = 0.01$$

Solution is around 6.5.

3.12

$X$ is Binomial($n,p$) and $Y$ is negative binomial($r,p$) (zero based, $Y$ counts number of failures).

$$F_{X}(r-1) = P( r-1 n )$$

$$= 1-P( r n )$$

$$= 1-P( r n )$$

$$= 1-P( r \le n-r )$$

$$= 1-F_{Y}(n-r)$$