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? 0.5625 0.018 0.032 0.15625 0.005

Answer 1

The probability that Mary packed the dinner is 0.15625

Further explanation

The probability of an event is defined as the possibility of an event occurring against sample space.

Let us tackle the problem.

This problem is about Conditional Probability.

in general, the formula we will use is as follows :

$\large {\boxed {P ( A | B ) = P ( A \bigcap B ) \div P ( B )} }$

Let:

The probability of John that packs the dinners without salt packet is P(J)

The probability of Mary that packs the dinners without salt packet is P(M)

The probability of Sue that packs the dinners without salt packet is P(S)

Then :

$P(J) = 45\% \times 4\% = 0.018$

$P(M) = 25\% \times 2\% = 0.005$

$P(S) = 30\% \times 3\% = 0.009$

If I find there is not salt in my purchased dinner, then the probability that Mary packed my dinner is :

$Probability = \frac{P(M)}{P(J) + P(M) + P(S)}$

$Probability = \frac{0.005}{0.018 + 0.005 + 0.009}$

$Probability = \frac{0.005}{0.032}$

$\large {\boxed {Probability = 0.15625} }$

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Probability , Sample , Space , Six , Dice , Die , Workers , Fast Food Restaurant , Salt Packet , Dinner

Answer 2

The probability of that Mary packed for dinner and the dinner contains no salt is $\boxed{0.15625}.$

Further explanation:

The probability can be obtained as the ratio of favorable number of outcomes to the total number of outcomes.

$\boxed{{\text{Probability}} = \frac{{{\text{Favorable number of outcome}}}}{{{\text{Total number of outcomes}}}}}$

Given:

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.

Explanation:

The probability of John packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{John}}} \right)&= 45\%\\&= 0.45\\\end{aligned}$

The probability of Mary packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{Mary}}} \right) &= 25\% \\&= 0.25\\\end{aligned}$

The probability of Sue packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{Sue}}} \right) &= 1 - 0.45 - 25\\&= 1 - 0.70\\&= 0.30\\\end{aligned}$

The probability of no salt packet is packed given that the John packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{John packs}}}}} \right) = 0.04$

The probability of no salt packet is packed given that the Mary packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Mary packs}}}}} \right) = 0.02$

The probability of no salt packet is packed given that the Sue packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Sue packs}}}}} \right) = 0.03$

The probability of that Mary packed for dinner and the dinner contains no salt can be expressed as follows,

${\text{P}}\left( {\dfrac{{{\text{Mary purchased}}}}{{{\text{No salt}}}}} \right) = \frac{{P\left( {{\text{Mary}}} \right) \times P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Mary}}}}} \right)}}{{P\left( {{\text{John}}} \right) \times P\left( {\dfrac{{{\text{No salt}}}}{{{\text{John}}}}} \right) + P\left( {{\text{Mary}}} \right) \times P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Mary}}}}} \right) + P\left( {{\text{Sue}}} \right) \times P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Sue}}}}} \right)}}$

The probability can be obtained as follows,

$\begin{aligned}{\text{P}}\left( {\frac{{{\text{Mary purchased}}}}{{{\text{No salt}}}}} \right) &= \frac{{0.25 \times 0.02}}{{0.45 \times 0.04 + 0.25 \times 0.02 + 0.30 \times 0.03}}\\&= \frac{{0.005}}{{0.018 + 0.005 + 0.009}}\\&= \frac{{0.005}}{{0.032}}\\&= 0.015625\\\end{aligned}$

The probability of that Mary packed for dinner and the dinner contains no salt is $\boxed{0.15625}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Three worker, fast food, restaurant pack, chicken dinners, John packs, 45% of the dinners, Mary, 25% of dinner, Sue packs remaining, 4% so not include salt, salt packet, salt is omitted, dinner, 2%, 3%, purchased dinner, probability, 0.5625, 0.018, 0.032, 0.15625, 0.005.