9 June, 00:30

# Three workers at a fast food restaurant pack the take-out chicken dinners. John packs 45% of the dinners, Mary packs 25% of the dinners and Sue packs the remaining dinners. Of the dinners John packs 4% do not include a salt packet. If Mary packs the dinner 2% of the time the salt is omitted. Lastly, 3% of the dinners do not include salt if Sue does the packing. If you find there is not salt in your purchased dinner what is the probability that Mary packed your dinner? 0.5625 0.018 0.032 0.15625 0.005

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1. 9 June, 04:18
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The probability of:

John packed your dinner: P (John) =.45

Mary packed your dinner: P (Mary) =.25

Sue packed your dinner: P (Sue) =.30

John packed, no salt packs: P (No salt | John) =.04

Mary packed, no salt packs: P (No salt | Mary) =.02

Sue packed, no salt packs: P (No salt | Sue) =.03

Total proportion of no salt pack is:

P (No Salt| John) * P (John) + P (No Salt | Mary) * P (Mary) + P (No Salt | Sue) * P (Sue)

= 0.04*0.45 + 0.02*.25 + 0.03*.30

= 0.0320

The probability of finding no salt in your purchased dinner, if Mary packed your dinner is:

P (Mary | No salt) = P (No Salt | Mary) * P (Mary) divided by total fraction of no salt in your dinner

=0.02*.25 / 0.0320

=0.15625