Three workers at a fast food restaurant pack the take-out chicken dinners. John packs 45% of the dinners, Mary packs 25% of the

Question

7 days ago

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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

Mathematics

2 answers:

Inessa [9K] · Answer 1 · 2026-03-21T01:27:13+03:00

The likelihood that Mary was the one who packed the dinner stands at 0.15625. To clarify, the probability of an event is defined as the chance that it will occur relative to the total possible outcomes. Let's analyze the problem at hand, which pertains to Conditional Probability. The formula we use generally is as follows: Let: The chance of John packing dinners without a salt packet is denoted as P(J), for Mary, it is P(M), and for Sue, it is P(S). Hence, if there's no salt in the dinner I purchased, the probability that Mary did the packing is calculated as follows.

Svet_ta [9.5K] · Answer 2 · 2026-03-21T01:28:42+03:00

The probability that Mary packed the dinner and that it lacks salt is determined by the favorable outcomes ratio to the total outcomes available. Given that John packs a percentage of the meals, Mary packs 25%, and Sue prepares the rest. Of the dinners John handles, 4% come without a salt packet; if Mary is packing, it’s 2% of the time that salt is left out. Lastly, for dinners packed by Sue, it’s 3% that don't have any salt. To elucidate further, we can express these probabilities accordingly.