Question

Hello. I have no idea how to solve this question below about Probability. I would appreciate if you show how to solve it step-by-step. Thank you so much.

Alonzo, Bob, and Casper work bussing tables at a restaurant. Alonzo has a 70% chance, Bob has a 10% chance, and Casper has a 20% chance of bussing tables in the middle area of the restaurant. If Alonzo is bussing tables, he has a 6% chance of breaking a dish. If Bob is bussing tables, he has a 2% chance of breaking a dish. Finally, if Casper is bussing tables, he has a 3% chance of breaking a dish. If there is a broken dish in the middle of the restaurant, what is the probability it was broken by Casper?

Answer 1

Alonzos will be making 45 percent

Answer 2

The probability the dish was broken by Casper if he bussed tables is;

P(Casper bussing tables and breaking dish) = 0.6%

We are given the three probabilities each for Alonzo, Bob and Casper for bussing tables and also in case any dish should break.

P(Alonzo bussing tables) = 70% = 0.7

P(Bob bussing tables) = 10% = 0.1

P(Casper bussing tables) = 20% = 0.2

P(Alonzo breaking a dish) = 6% = 0.06

P(Bob breaking a dish)  = 2% = 0.02

P(Casper breaking a dish)  = 3% = 0.03

This is all about independent events as the probability of breaking the dishes does not increase or decrease with the bussing of the tables.

Thus; P(A and B) = P(A) + P(B)

Thus, if a dish is broken, the probability that it was broken by Casper is;

P(Casper bussing tables and breaking dish) = P(Casper bussing tables) × P(Casper breaking a dish)

P(Casper bussing tables and breaking dish) = 0.2 × 0.03

P(Casper bussing tables and breaking dish) = 0.006 or 0.6%

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