Question

Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A, be the event that the receiver functions properly throughout the warranty period, A, be the event that the speakers function properly throughout the warranty period, and A, be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A) = 0.91, P(A) = 0.92, and P(A3) = 0.90. (Round your answers to four decimal places.) (a) What is the probability that all three components function properly throughout the warranty period? 0.7535 (b) What is the probability that at least one component needs service during the arranty period? 02465 (c) What is the probability that all three components need service during the warranty period? 0.0072 X (d) What is the probability that only the receiver needs service during the warranty period? 0.0655 (e) What is the probability that exactly one of the three components needs service during the warranty period? 0 2238

Answer 1

(a)

Since components are independent from each other so the required probability is

$P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})P(A_{2})P(A_{3})=0.91\cdot 0.92\cdot 0.90=0.75348\approx 0.7535$

(b)

The probability that at least one component need service using Demorgan's law is

PA UA UA) = 1-PANA NA3) = P(A1P(A2) P(A3) = 1-0.7535 = 0.2465

(c)

The probability that all three components need service is

$[1-P(A_{1})][1-P(A_{2})][1-P(A_{3})]=[1-0.91][1-0.92][1-0.90]=0.00072$

Answer: 0.0007

(d)

The probability only the receiver needs service during the warranty period is

$[1-P(A_{1})][P(A_{2})][P(A_{3})]=[1-0.91][0.92][0.90]=0.07452$

Answer: 0.0745

(e)

$[1-P(A_{1})][P(A_{2})][P(A_{3})]+[P(A_{1})][1-P(A_{2})][P(A_{3})]+[P(A_{1})][P(A_{2})][1-P(A_{3})]=[1-0.91][0.92][0.90]+[0.91][1-0.92][0.90]+[0.91][0.92][1-0.90]=0.22376$

Answer: 0.2238