A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, turntable). A purchaser is offered a choice of manufacturer for each component:
Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood
Compact disc player: Onkyo, Pioneer, Sony, Technics
Speakers: Boston, Infinity, Polk
Turntable: Onkyo, Sony, Teac, TechnicsA switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions:
a. In how many ways can one component of each type be selected?
b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony?
c. In how many ways can components be selected if none is to be Sony?
d. In how many ways can a selection be made if at least one Sony component is to be included?
e. If someone flips switches on the selection in a completely random fashion, what is the probability that the
system selected contains at least one Sony component?Exactly one Sony component?

[tex] \left(\begin{array}{ccc}5\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) [/tex]

[tex] = 5 * 4 * 3 * 4 = 240 ways [/tex]

b) If both the receiver and compact disk are to be sony.

In the receiver, the purchaser was offered 1 Sony, also in the CD(compact disk) player the purchaser was offered 1 Sony.

Thus, the number of ways components can be selected if both receiver and player are to be Sony is:

[tex] \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) [/tex]

[tex] = 1 * 1 * 3 * 4 = 12 ways [/tex]

c) If none is to be Sony.

Let's exclude Sony from each component.

Receiver has 1 sony = 5 - 1 = 4

CD player has 1 Sony = 4 - 1 = 3

Speakers had 0 sony = 3 - 0 = 3

Turntable has 1 sony = 4 - 1 = 3

Therefore, the number of ways can be selected if none is to be sony:

[tex] \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) [/tex]

[tex] = 4 * 3 * 3 * 3 = 108 ways [/tex]

d) If at least one sony is to be included.

Number of ways can a selection be made if at least one Sony component is to be included =

Total possible selections - possible selections without Sony

= 240 - 108

= 132 ways

e) If someone flips switches on the selection in a completely random fashion.

i) Probability of selecting at least one Sony component=

Possible selections with at least one sony / Total number of possible selections

[tex] \frac{132}{240} = 0.55 [/tex]

ii) Probability of selecting exactly one sony component =

Possible selections with exactly one sony / Total number of possible selections.

[tex] \frac{\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right)}{240} [/tex]

[tex] = \frac{(1*3*3*3)+(4*1*3*3)+(4*3*3*1)}{240} [/tex]

[tex] \frac{27 + 36 + 36}{240} = \frac{99}{240} = 0.4125 [/tex]