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3 years ago

I need help again please

Mathematics

1 answer:

Arisa [49]3 years ago

5 0

I hope this helps you

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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

Gekata [30.6K]

The probability that Mary packed the dinner is 0.15625

<h3>Further explanation</h3>

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 :

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

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

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

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

<h3>Learn more</h3>

Different Birthdays : brainly.com/question/7567074

Dependent or Independent Events : brainly.com/question/12029535

Mutually exclusive : brainly.com/question/3464581

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Probability

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

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