Question

1. ashley, bob, claire, and daniel are among 13 students who entered a lottery to win a free vacation to paris. only 4 people will be chosen at random. what is the probability that ashley, bob, claire, and daniel will be chosen? (lesson 19.3) (1 point)

a. $\frac{1}{17160}$
b. $\frac{4}{715}$
c. $\frac{1}{440}$
d. $\frac{1}{715}$

1. a six - sided die is rolled 4 times and the results are recorded. answer the following. (lesson 19.3) (1 point total)

a) how many outcomes are in the sample space, $n(s)$?
b) how many outcomes are there of rolling a 6 exactly 2 times, $n(a)$?
c) what is the probability of rolling a 6 exactly 2 times? $p(a)=\frac{n(a)}{n(s)}=$

1. you just got free tickets for a boat ride, and you can only bring along 2 friends. unfortunately, you have 6 friends who want to come along. how many different groups of friends could you take with you? (lesson 19.3) (1 point)

a. 720
b. 30
c. 15

Explanation:

Question 7

a)

Step1: Use the multiplication principle

Each roll of the die has 6 possible outcomes. For 4 rolls, the total number of outcomes in the sample space \(n(S)\) is \(6\times6\times6\times6\).
$$n(S)=6^4 = 1296$$

Step1: Use the binomial - coefficient formula

We use the formula \(C(n,k)=\frac{n!}{k!(n - k)!}\) (where \(n = 4\) rolls and \(k=2\) successes (rolling a 6)) and then consider the non - 6 outcomes. The number of ways to choose 2 positions out of 4 for the 6s is \(C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3\times2!}{2!×2!}=6\). The non - 6 outcomes (there are \(4 - 2=2\) non - 6 rolls) have \(5\) possibilities each. So \(n(A)=C(4,2)\times5^2\).

Step2: Calculate the value

\(C(4,2)\times5^2=6\times25 = 150\)

Step1: Use the probability formula

The probability formula is \(P(A)=\frac{n(A)}{n(S)}\). We know \(n(A) = 150\) and \(n(S)=1296\).
$$P(A)=\frac{150}{1296}=\frac{25}{216}$$

Answer:

\(1296\)

b)