Question

Cell phone handoff behavior. Refer to the Journal of Engineering, Computing and Architecture (Vol. 3., 2009) study of cell phone handoff behavior, Exercise 3.47 (p. 183). Recall that a “handoff” describes the process of a cell phone moving from one base channel (identified by a color code) to another. During a particular driving trip, a cell phone changed channels (color codes) 85 times. Color code “b” was accessed 40 times on the trip. You randomly select 7 of the 85 handoffs. How likely is it that the cell phone accessed color code “b” only twice for these 7 handoffs?

Short answer

19.31% is not most likely to have the cell phone accessed colorcode b only twice.

Step-by-step solution

Given information

Refer to the Journal of Engineering Computing and Architecture (Vol. 3., 2009), on a particular driving trip, a cell phone changes the channels (color codes) 85 times, i.e., N is 85, color code b was accessed 40 times on the trip, i.e., r is 40, and select 7 of the 85 handoffs are randomly selected, that is n is 7.

Calculate the probability that the cell phone accessed color code b only twice

The random variable x is the number of times the cell phone accessed color code b

Here, x follows a hypergeometric distribution with$N=85, n=7 and r=40$

The probability mass function of x is given by,

role="math" localid="1659709074655" $P\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

So,

$x=2$

$\begin{array}{rcl}P\left(x=2\right)& =& \frac{\left(\begin{array}{c}40\\ 2\end{array}\right)\left(\begin{array}{c}85-40\\ 7-2\end{array}\right)}{\left(\begin{array}{c}85\\ 7\end{array}\right)}\\ & =& \frac{\left(\begin{array}{c}40\\ 2\end{array}\right)\left(\begin{array}{c}45\\ 5\end{array}\right)}{\left(\begin{array}{c}85\\ 7\end{array}\right)}\\ & =& \frac{780\times 1221759}{4935847320}\\ & =& \frac{952972020}{4935847320}\\ & & \\ & & \end{array}$

$$\begin{gathered} =0.193071616 \\ \approx 0.1931 \end{gathered}$$

$P\left(x=2\right)=0.1931$

Thus, the probability that the cell phone accessed color code b is only 0.1931.

Therefore, it is not very likely that the cell phone accessed colorcode b only twice because there is only a 19.31% chance of occurring.