Question

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?

Answer 1

Answer:0.5625

Step-by-step explanation:

If John packed the dinner

probability of salt less dinner=$\frac{45}{100} \times \frac{4}{100}$=0.018

If Mary packed the dinner

probability  of salt-less dinner =$\frac{25}{100} \times \frac{2}{100}$=0.005

If sue packed the dinner

probability  of salt-less dinner =$\frac{30}{100} \times \frac{3}{100}$=0.009

Total proportion of salt-less dinner=0.032

probability that dinner is salt-less and mary packed it=$\frac{0.018}{0.032}$=0.5625