Calculating Probability of Reparation Costs for a Device with Fragile Pieces

[penguin007]

Hi everyone,
I just would like to be reassured about simple probability question:

Homework Statement

A device is composed by two fragile pieces
during the warranty period the probability that:
-none is broken is 0,25
-both are broken is 0,25
-only one is broken is 0,25
If one piece is broken then the probability that it costs
-100$ is 1/2
-300$ is 1/3
-600$ is 1/6
What is the probability that the reparation costs 900$ ?



3. The attempt of a solution


- P = ( probability that both pieces are broken that the first costs 600$, the second 300$ and the first costs 300$ and the second 600$ )
= ( 0,25 * 1/3 * 1/6 ) * 2

Do you agree with me ?


Thanks!

 

[mighty2000]

Hello!

Thank you for your question. Based on the information given, your solution appears to be correct. The probability of both pieces being broken and having a repair cost of 600$ and 300$ is (0.25 * 1/3 * 1/6) = 1/144, and since there are two possible combinations (600$ and 300$ or 300$ and 600$), we multiply by 2 to get a total probability of 2/144, which simplifies to 1/72. This means that there is a 1/72 chance that the reparation cost will be 900$. I hope this helps to reassure you. Let me know if you have any further questions. Good luck with your homework!