Chapter 12.3, Problem 23HP

To determine

To calculate: the probability that a student is a freshman and he or she does have a myroom page.

Answer to Problem 23HP

The required probability is $43/66$

Explanation of Solution

Given information:

Mr. Woodruff gives a pop quiz at the beginning of class 15% of the time.

Calculation:

From the given information we have,

  $\begin{array}{r}\hfill P(\text{MyRoom page}\mid \text{freshman})=\frac{43}{55}\\ \hfill P(\text{not MyRoom page}\mid \text{sophomore})=\frac{4}{27}\end{array}$

Since total number of students is 82 so number of students that has MyRoom page and freshman is

  $n$ (MyRoom page and freshman)=43

By the complement rule we have,

P (MyRoom page\sophomore) =1 $-$ P (not MyRoom page\sophomore)

  $\begin{array}{l}=1-\frac{4}{27}\hfill \\ =\frac{23}{27}\hfill \end{array}$

Since total number of students is 82 so number of students that has MyRoom page and sophomore is

  $n$ (MyRoom page and sophomore)=23

Therefore, total number of students having MyRoom page is,

  $n$ (MyRoom page)= $n$ ( MyRoom page and freshman)+ $n$ (MyRoom page and sophomore)

  $\begin{array}{l}=43+23\\ =66\end{array}$

So the probability that a student is a freshman, given that he or she does have a MyRoom page is,

  $\begin{array}{cc}P(\text{freshman}\mid \text{MyRoom page})& =\frac{n(\text{MyRoom page and freshman})}{n(\text{MyRoom page})}\\ & =\frac{43}{66}\end{array}$

Hence, the required probability is $43/66$