Four workers at a fast food restaurant pack the take-out chicken dinners. john packs 45% of the dinners but fails to include a salt packet 4% of the time. mary packs 25% of the dinners but omits the salt 2% of the time. sue packs 30% of the dinners but fails to include the salt 3% of the time. you have purchased a dinner and there is no salt.
a. find the probability that john packed your dinner. b. find the probability that mary packed your dinner.

Step-by-step explanation:

This scenario is a perfect application of Bayes' Theorem. We have:

P(John) = .45

P(Mary) = .25

P(Sue) = .30

P(No Salt|John) = .04

P(No Salt|Mary) = .02

P(No Salt|Sue) = .03

In part a, we want to know P(John|No Salt). Applying Bayes:

P(John|No Salt) = P(No Salt|John)P(John) / P(No Salt|John)P(John)+P(No Salt|Mary)P(Mary)+P(No Salt|Sue)P(Sue)

= (.04)(.45) / (.04)(.45) + (.02)(.25) + (.03)(.30)

= 0.018 / (0.018 + 0.005 + 0.009)

= 0.5625

In part b, we want to know P(Mary|No Salt). Again, applying Bayes' in the same way:

P(Mary|No Salt) = P(No Salt|Mary)P(Mary) / P(No Salt|John)P(John)+P(No Salt|Mary)P(Mary)+P(No Salt|Sue)P(Sue)

= (.02)(.25) / (.04)(45) + (.02)(.25) + (.03)(.30)

= 0.5 / (1.8 + 0.5 + 0.9)

= 0.15625

the balance is found at the link

step-by-step explanation:

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent