Mathematics, 10.11.2019 03:31 kat1357

According to the u. s. bureau of labor statistics, 75% of the women 25 through 49 years of age participate in the labor force. suppose 78% of the women in that age group are married. suppose also that 61% of women 25 through 49 years of age are married and are participating in the labor force. (round your answers to 2 decimal places.)a. what is the probability that a randomly selected woman in that age group is married or is participating in the labor force? p = . what is the probability that a randomly selected woman in that age group is married or is participating in the labor force but not both? p = . what is the probability that a randomly selected woman in that age group is neither married nor participating in the labor force? p =

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a) There is a 92% probability that a randomly selected woman in that age group is married or is participating in the labor force.

b) There is a 31% probability that a randomly selected woman in that age group is married or is participating in the labor force but not both.

c) This means that there is an 8% probability that a randomly selected woman in that age group is neither married nor participating in the labor force.

Step-by-step explanation:

We can solve this problem building the Venn's diagram of these probabilities.

I am going to say that

The set A are those women who participate in the work force.

The set B are those women who are married.

We have that:

$A = a + (A \cap B)$

In which a represents those that participate in the work force but are not married and $(A \cap B)$ are those who are both on the work force and married.

By the same logic, we have that:

$B = b + (A \cap B)$

We start finding the values from the intersection of these sets:

Suppose also that 61% of women 25 through 49 years of age are married and are participating in the labor force.

This means that $A \cap B = 0.61$

Suppose 78% of the women in that age group are married.

This means that B = 0.78

$B = b + (A \cap B)$

0.78 = b + 0.61

b = 0.17

75% of the women 25 through 49 years of age participate in the labor force.

This means that A = 0.75

$A = a + (A \cap B)$

0.75 = a + 0.61

a = 0.14

a. What is the probability that a randomly selected woman in that age group is married or is participating in the labor force

This is

$P = a + b + (A \cap B) = 0.14 + 0.17 + 0.61 = 0.92$

There is a 92% probability that a randomly selected woman in that age group is married or is participating in the labor force.

b. What is the probability that a randomly selected woman in that age group is married or is participating in the labor force but not both?

This is

P = a + b

P = 0.14 + 0.17

P = 0.31

There is a 31% probability that a randomly selected woman in that age group is married or is participating in the labor force but not both.

c. What is the probability that a randomly selected woman in that age group is neither married nor participating in the labor force?

From a), we found that there is a 92% probability that a randomly selected woman in that age group is married or is participating in the labor force.

This means that there is an 8% probability that a randomly selected woman in that age group is neither married nor participating in the labor force.

Answer from: Quest

Answer for 2: his profit would be $2.88 explanation: you have to first multiply 72 and 0.12 together to get $8.64 once you do that you know how much money the shopkeeper made from selling the candy bars individually. so you have to do 8.64-5.76=2.88 hope that ; )answer for 3: yeah i’m sorry i have no idea what number 3 is. i’m sorry i couldn’t with that.
I'm going to add u to plsss answer questions 2 n 3

I'm going to add u to plsss answer questions 2 n 3

Answer from: Quest

jk it's 20

step-by-step explanation:

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