A computer center has three printers: A, B, and C. Programs are routed to the first available printer. The printers print at different speeds and the probability that a program is routed to printers A, B, and C are equal to 0.5, 0.3, and 0.2, respectively. Occasionally, printers A, B, and C, jam and destroy printouts with the probabilities 0.04, 0.05, and 0.03, respectively. Your program is destroyed when a printer jams.

What is the probability that printer A is involved given that there is a jam?

Give your answer to 4 decimal places

Question

A computer center has three printers: A, B, and C. Programs are routed to the first available printer. The printers print at different speeds and the probability that a program is routed to printers A, B, and C are equal to 0.5, 0.3, and 0.2, respectively. Occasionally, printers A, B, and C, jam and destroy printouts with the probabilities 0.04, 0.05, and 0.03, respectively. Your program is destroyed when a printer jams.

What is the probability that printer A is involved given that there is a jam?

Give your answer to 4 decimal places

anonymous · Answer

$$p = (0.5)(0.04) / [ (0.5)(0.04) + (0.3)(0.05) + (0.2)(0.03)]$$

anonymous · Answer

This seems to be a question on conditional probability, so I tried to set it up using Baye's Theorem..Can you explain your approach please. I actually already had the "answer". Need to understand the method. Thanks.

anonymous · Answer

ok. I just used the idea behind bayes theorem. In my numerator, I have the chance of A getting jammed. In the denominator, I have all the three A,B,C chance of getting jammed.
Can you get it?