Dan's diner employs three dishwashers. al washes 60 % of the dishes and breaks only 1 % of those he handles. betty and chuck each wash 20 % of the dishes, and betty breaks only 1 % of hers, but chuck breaks 3 % of the dishes he washes. you go to dan's for supper one night and hear a dish break at the sink. what's the probability that chuck is on the job?

0.428

Step-by-step explanation:

Given:

P(Al) = 0.6

P(break | Al) = 0.01

P(Betty) = 0.2

P(break | Betty) = 0.01

P(Chuck) = 0.2

P(break | Chuck) = 0.03

Now,

The probability that the chuck is on the job will be calculated as:

P(Chuck | break) =  .............(1)

Now,

P(break) is the probability that the plate will break

= P(break | Al) × P(Al) + P(break | Betty) × P(Betty) + P(break | Chuck) × P(Chuck)

= 0.01 × 0.6 + 0.01 × 0.2 + 0.03 × 0.2

= 0.006 + 0.002 + 0.006

= 0.014

substituting the value in (1), we get

P(Chuck | break) =  

P(Chuck | break) = 0.428

idk

step-by-step explanation:

so then what are you asking?

step-by-step explanation: