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 directly into the old mainframe. Historically, we know the following probabilities:
i. P(light pen will fail) = 0.025
ii. P(PC key board will fail) = 0.15
iii. P(light pen and PC keyboard will fail) = 0.005
iv. P(Main frame will fail) =0.25
Data can be entered into the PC only if both the light pen and keyboard are functioning
a. What is the probability that the employee can use the PC to enter data?
b. What is the probability that either the PC fails or the mainframe fails? (assume they cannot both fail at the
same time)

Answer 1

Step-by-step explanation:

here,

Given,P(l)=0.25

p(pc)=0.15

P(l^pc)=0.005

P(M)=0.25

condition:data can be entered only if both the light pen and keyboard are functioning

solution,

p(light and pc functioning)=1-0.005=0.095

a.

p(employee can use pc to enter data)=(1-0.15)*0.095=0.08075.

b.

P(either pc fails or the mainframe fails)=p(pc)UP(M)

=0.15+0.25=0.4

Answer 2

Answer:

Step-by-step explanation:

P(L) = 0.025, P(K) = 0.15, P(L intersection K) = 0.005, P(M) = 0.25

a. P(L union K) = 0.025+0.15-0.005 = 0.17

Probability that employee can use PC to enter data,

P( complement of L union K) = 1-0.17 = 0.83

b. Since they can't both fail at same times, Probaility = 0.025 + 0.15 = 0.175