Question

Three 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.

Answer 1

Solution from the given Information, Here, his scenario is a perfect application Of Baye's Theroem. we have P(John) = 0.45 l Good and p(mary) = 0.25 P(Sue) = 0.30 plno salt) John) -0.04 P( No salt) Mary) = 0.02 P( No salt) sue) = 0.03 @ Here, we want to know p (John) Nosalt) ; Applying Baye's; » P(Johol Nosalt) = P(No salt/John) p (John) / plNosalt/ John), (John) + p (No salt/mary) (masy) + Pl Nosalt/ sue) p (sue). > (0.04)(0:43) / (0:4)[0.45)+10:03) (0-25) + 10-03)(0.30) => 0.018 / 10.018 + 0.005 +0.009) - 10.5685

6 Here We want to know, pl Mary/ No salt); Applying Baye's p(Maryl No salt) = P(No salt / Mary) P(mary /p/Nosalt/ Johop (John) + p (Nosalt/masy) p (many) + Plno salt/ sue) p (sue) 10.02) (0-25)) (0.04) (45)+10.03)10.25)+ 10.03) (0.30) = 0.5 111.870.5+0.9) 10.15625
Thank you.