EX 2.29

• An electronics store sells three different brands of DVD players.
• Of its DVD player sales,
  • $50\%$ are brand 1 (the least expensive),
  • $30\%$ are brand 2,
  • $20\%$ are brand 3 .
• Each manufacturer offers a 1-year warranty on parts and labor.
• It is known that $25\%$ of brand 1’s DVD players require warranty repair work,
  • whereas the corresponding percentages for brands 2 and 3 are $20\%$ and $10\%$ , respectively.

Question

1. What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty?
2. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty?
3. If a customer returns to the store with a DVD player that needs warranty repair work,

• what is the probability that it is
  • a brand 1 DVD player?
  • A brand 2 DVD player?
  • A brand 3 DVD player?

• The first stage of the problem involves a customer selecting one of the three brands of DVD player.

• Let

  • $A_i=\{$ brand $i$ is purchased $\}$ , for $i=1$, $2$ , and $3$

• Then

  • $P\left(A_1\right)=.50$,
  • $P\left(A_2\right)=.30$ ,
  • $P\left(A_3\right)=.20$ .

• Once a brand of DVD player is selected,

  • the second stage involves observing whether the selected DVD player needs warranty repair.

• With

  • $B=\{$ needs repair $\}$
  • $B^{\prime }=\{$ doesn’t need repair $\}$ ,

• the given information implies that

  • $P\left(B\mid A_1\right)=.25$,
  • $P\left(B\mid A_2\right)=.20$ ,
  • $P\left(B\mid A_3\right)=.10$ .

• The tree diagram representing this experimental situation is shown in Figure 2.10.

Figure 2.10 Tree diagram for Example 2.29

• The initial branches correspond to different brands of DVD players;
• there are two second-generation branches emanating from the tip of each initial branch,
  • one for “needs repair” and
  • the other for “doesn’t need repair.”
• The probability $P\left(A_i\right)$ appears on the $i$ th initial branch,
  • whereas the conditional probabilities $P\left(B\mid A_i\right)$ and $P\left(B^{\prime }\mid A_i\right)$ appear on the second-generation branches.
• To the right of each second-generation branch corresponding to the occurrence of $B$ ,
  • we display the product of probabilities on the branches leading out to that point.
• This is simply the multiplication rule in action.
• The answer to the question posed in 1 is thus $P\left(A_1\cap B\right)=P\left(B\mid A_1\right)\cdot P\left(A_1\right)=.125$
• The answer to question 2 is

$$\begin{array}{rl}P\left(B\right)& =P\left[\text{(brand 1 and repair) or (brand 2 and repair) or (brand 3 and repair)}\right]\\ & =P\left(A_1\cap B\right)+P\left(A_2\cap B\right)+P\left(A_3\cap B\right)\\ & =.125+.060+.020=.205\end{array}$$

Finally,

$$P\left(A_1\mid B\right)=\frac{P\left(A_1\cap B\right)}{P\left(B\right)}=\frac{.125}{.205}=.61$$ $$P\left(A_2\mid B\right)=\frac{P\left(A_2\cap B\right)}{P\left(B\right)}=\frac{.060}{.205}=.29$$

and

$$P\left(A_3\mid B\right)=1-P\left(A_1\mid B\right)-P\left(A_2\mid B\right)=.10$$

• The initial or prior probability of brand 1 is .50 .
• Once it is known that the selected DVD player needed repair,
  • the posterior probability of brand 1 increases to .61.
  • This is because brand 1 DVD players are more likely to need warranty repair than are the other brands.
• The posterior probability of brand 3 is $P\left(A_3\mid B\right)=.10$ ,
  • which is much less than the prior probability $P\left(A_3\right)=.20$ .