# Answer to Question #5219 in Statistics and Probability for cash

Question #5219

Three stores have 8, 12, and 15 employees of whom 3, 8, and 7, respectively, are women.

(i) A store is chosen at random and from that store an employee is chosen at random. If this employee is a woman, what is the probability she came from the store with 12 employees?

(ii) If a second employee is chosen from the same store in (i), what is the probability that a woman will be chosen? Assume that the employee chosen in (i) was a woman and we don’t know which store she came from.

[7 marks]

(i) A store is chosen at random and from that store an employee is chosen at random. If this employee is a woman, what is the probability she came from the store with 12 employees?

(ii) If a second employee is chosen from the same store in (i), what is the probability that a woman will be chosen? Assume that the employee chosen in (i) was a woman and we don’t know which store she came from.

[7 marks]

Expert's answer

Consider the events:

A - the store with 8 employees is chosen,

B - the

store with 12 employees is chosen,

A - the store with 15 employees is

chosen,

W - employee is a woman.

i)

We need to find the conditional

probability

P(B|W) = |using Bayes' theorem| = (P(W|B)*P(B)) / (P(W|A) +

P(W|B) + P(W|C));

P(B|W) = (8/12*1/3) / (3/8 + 8/12 + 7/15) =

80/543.

ii)

Let's first calculate

P(A|W) and P(C|W) like in part

i):

P(A|W) = (3/8*1/3) / (3/8 + 8/12 + 7/15) = 45/543.

P(C|W) = (7/15*1/3)

/ (3/8 + 8/12 + 7/15) = 56/543.

Using the law of total probability, we

get:

P(W) = P(W|A)*P{the first woman was from A} + P(W|B)*P{the first woman

was from B} +

P(W|C)*P{the first woman was from C} = 2/8*45/543 +

7/12*80/543 + 6/15*56/543 = 1/543*(45/4 + 140/3 + 112/5) = 4819/(543*60) =

0.148...

A - the store with 8 employees is chosen,

B - the

store with 12 employees is chosen,

A - the store with 15 employees is

chosen,

W - employee is a woman.

i)

We need to find the conditional

probability

P(B|W) = |using Bayes' theorem| = (P(W|B)*P(B)) / (P(W|A) +

P(W|B) + P(W|C));

P(B|W) = (8/12*1/3) / (3/8 + 8/12 + 7/15) =

80/543.

ii)

Let's first calculate

P(A|W) and P(C|W) like in part

i):

P(A|W) = (3/8*1/3) / (3/8 + 8/12 + 7/15) = 45/543.

P(C|W) = (7/15*1/3)

/ (3/8 + 8/12 + 7/15) = 56/543.

Using the law of total probability, we

get:

P(W) = P(W|A)*P{the first woman was from A} + P(W|B)*P{the first woman

was from B} +

P(W|C)*P{the first woman was from C} = 2/8*45/543 +

7/12*80/543 + 6/15*56/543 = 1/543*(45/4 + 140/3 + 112/5) = 4819/(543*60) =

0.148...

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## Comments

Assignment Expert13.07.15, 19:49Dear tope.

Your comment is not related to question 5219, but it is the same as question 53385, which is in progress. Please wait for updates.

tope13.07.15, 14:38The following data show systolic blood pressure levels (mm Hg) of a random sample of six

patients undergoing a particular drug therapy for hypertension.

182 179 154 161 170 151

Can we conclude, on the basis of these data, that the population mean is greater than 165?

Hint: Use an appropriate parametric test, at the 5% significance level.

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