Question

1. there are 52 cards in a standard deck. 12 of those cards are face cards (4 jacks, 4 queens, and 4 kings). deck is shuffled and a group of 4 cards is dealt. the probability that all 4 cards are face cards is ______. round your answer to the nearest thousandth. (lesson 19.3) (1 point) 10. an office has 7 male employees and 8 female employees. the manager randomly chooses 2 employees to attend a football game. what is the probability that the manager chooses 2 female employees? (lesson 19.3) (1 point) a. $\frac{1}{15}$ b. $\frac{2}{15}$ c. $\frac{4}{15}$ d. $\frac{8}{15}$ 11. use the following information for problems a - b. the table shows the results of a school lunch survey. in the survey, students were asked whether they have chores and a curfew. (lesson 20.1) (1 point each) a) what is the probability that a student who has no curfew has no chores? a. $\frac{7}{80}$ b. $\frac{21}{66}$ c. $\frac{7}{20}$ d. $\frac{10}{11}$ b) what is the probability that a student who has no chores has a curfew? a. $\frac{3}{16}$ b. $\frac{1}{4}$ c. $\frac{11}{30}$ d. $\frac{15}{22}$

Explanation:

Step1: Calculate total number of ways to choose 2 employees

The total number of employees is \(7 + 8=15\). The number of ways to choose 2 employees out of 15 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 15\) and \(r=2\).

$$C(15,2)=\frac{15!}{2!(15 - 2)!}=\frac{15\times14}{2\times1}=105$$

Step2: Calculate number of ways to choose 2 female employees

The number of female employees is 8. The number of ways to choose 2 female employees out of 8 is \(C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28\)

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P=\frac{C(8,2)}{C(15,2)}=\frac{28}{105}=\frac{4}{15}\)

Step1: For part a)

We use the formula for conditional probability \(P(A|B)=\frac{n(A\cap B)}{n(B)}\). Here, \(A\) is “has no chores” and \(B\) is “has no curfew”. \(n(A\cap B) = 21\) (from the table) and \(n(B)=60\). So \(P=\frac{21}{60}=\frac{7}{20}\)

Step2: For part b)

We use the formula \(P(A|B)=\frac{n(A\cap B)}{n(B)}\). Here, \(A\) is “has a curfew” and \(B\) is “has no chores”. \(n(A\cap B) = 45\) and \(n(B)=66\). So \(P=\frac{45}{66}=\frac{15}{22}\)

Answer:

C. \(\frac{4}{15}\)

---