Answer to Question #284096 in Statistics and Probability for Savi

Question

"An employee at Infotech must enter product information into the computer. The employee may use a light pen that transmits the information to the PC along with the keyboard to issue Commands or fill out a bubble sheet and feed it to directly into the mainframe. Historically, we know the following probabilities: P(Light pen will fail) =0.025 P(PC keyboard will fail) =0.15 P(Light pen and PC keyboard will fail) =0.005 P(Mainframe will fail) =0.25 Data can be entered into the PC only if both the light pen and keyboard are functioning. Applying the concept of conditional probability , Solve the following (a) Chances of employee can use the PC to enter data? (b) Chance that either the PC fails or the mainframe fails? Assume they cannot both fail at the same time. "

Answer 1

Let $L$ denote the event "Light pen will fail".

Let $K$ denote the event "PC keyboard will fail".

Let $M$ denote the event "Mainframe will fail".

Given $P(L)=0.025, P(K)=0.15, P(L\cap K)=0.005,$

$P(M)=0.25$

(a)

$$P(\text{use the PC to enter data})=1-P(\text{PC fails)}$$

$$=1-P(L\cup K)$$

$$=1-(P(L)+P(K)-P(L\cap K))$$

$$=1-(0.025+0.15-0.005)$$

$$=1-0.17$$

$$=0.83$$

There is 83% chance of using the PC to enter data.

(b)

Assume that $P((\text{PC fails}) \cap M )=P((L\cup K) \cap M )=0.$

$$P((L\cup K)\cup M)=P(L\cup K)+P(M)$$

$$-P((L\cup K) \cap M )=$$

$$=0.17+0.25-0$$

$$=0.42$$

There is 42% chance that either the PC fails or the mainframe fails.