Answer to Question #170939 in Statistics and Probability for rohit

Question

A computer center has three printers A, B and C, which print at different speeds. Programmes are routed to the first available printer. The probability that the programmes are routed to the first available printer. The probability that the programmes are routed to the printers A, B and C are 0.6, 0.3 and 0.1 respectively. Occasionally, a printer will jam and destroy a printout. The probability that printers A, B and C will jam are 0.01, 0.05 and 0.04 respectively. Your programme is destroyed when a printer jams. What is the probability that printer A is involved?

Answer 1

Let E denote that printer is Jam-

Probability that program is routed to printer A is P(A)=0.6

Probability that program is routed to printer B is P(B)=0.3

Probability that program is routed to printer C is P(C)=0.1

Probability that printer A is Jam $P(\dfrac{A}{E})=0.01$

Probability that printer B is Jam $P(\dfrac{B}{E})=0.05$

Probability that printer C is Jam $P(\dfrac{C}{E})=0.04$

Probability that programme is destroyed when printer A jams-

$P(\dfrac{E}{A})=\dfrac{P(A).P(\dfrac{A}{E})}{P(A).P(\dfrac{A}{E})+P(B).P(\dfrac{B}{E})+P(C).P(\dfrac{C}{E})}$

$=\dfrac{0.6\times 0.01}{0.06\times 0.01+0.3\times 0.05+0.1\times 0.04}$

$=\dfrac{0.006}{0.006+0.015+0.004}=\dfrac{0.006}{0.025}=0.24$

Hence The required Probability is 0.24