Answer to Question #190563 in Algebra for Andrea

Question

1. Evaluate each expression

a. 8P5​b. 11C4

 

2. How many different four-letter passwords can be formed from the letters A, B, C, D, E, F and G if no repetition of letters is allowed?

 

3. An ice cream store sells two drinks in four sizes and has five flavor add ons. In how many ways can a customer order a drink?

 

 

4. An election ballot asks voters to select three city commissioners from a group of six candidates. In how many ways can this be done?

 

5. What is the multiplication rule for independent events?

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6. If P(A) = 0.25, P(B) = 0.60 and P(A and B) = 0.15, are the event A and B independent? Explain/show work. 

 

Answer 1

1.

we use the formula;

$nPr(n,r)=\frac{n!}{(n-r)!}$

$nCr(n,r)=\frac{n!}{(n-r)!\times r!}$

$a. 8P5​$

$nPr(8,5)=\frac{8!}{(8-5)!}$

$\frac{8\times 7\times6\times5\times4\times3!}{3!}$

$=6720$

$b. 11C4$

$nCr(11,4)=\frac{11!}{(11-4)!\times 4!}$

$\frac{11\times10\times9\times8\times7!}{7!\times4!}$

$=\frac{11\times10\times9\times8}{4!}$

$\frac{11\times10\times9\times8}{4\times3\times2\times1}$

$=330$

2.

Given 

A,B,C,D,E,F and G are 7 letters.

Formula:

$p(n,r)= \space^nP_r=\frac{n!}{(n-r)!}$

We have to choose 4 letters out of 7 letters and arrange them because here arrangement does matter ,so we will use permutation with n=7,r=4

$p(7,4)=\space ^7P_4=\frac{7!}{(7-4)!}$

$=\frac{7!}{3!}$

$\frac{7\times6\times5\times4\times3!}{3!}$

$7\times6\times5\times4$

$=840$

$p(7,4)=\space ^7P_4=840$

3.

From the provided information,

The ice cream store sells two drinks in four size and has five flavor add ons.

2 drinks can be chosen in 2 ways.

From the 4 sizes the one can be chosen in 4 ways.

From the 5 flavor the one can be chosen in 5 ways.

The required number of ways a customer can order the drink can be obtained as:

Number of ways to order a drink $=2\times4\times5=40$

there are 40 ways a customer can order for drinks

4.

To select 3 city commissioners

Total number of candidates is 6.

So, the total number of possibilities will be $n_{c_r}$ where n=6 and r=3

$n_{c_r}=\frac{n!}{(n-r)!\times r!}$

$=\frac{6!}{(6-3)!\times 3!}$

$=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times3!}$

$=5\times4$

$=20$

so, the selection of the three commissioners can be done in 20 ways

5.

  The multiplication rule for independent events associates the probabilities of two events to the probability in which they both occur. In case if we use the rule, then it is essential to have the probabilities of each of the independent events. By these events, the multiplication rule states that the probability that both events occur is found by multiplying the probabilities of each event.

Formula:

Consider events A and B and the probabilities of each by P(A) and P(B). If A and Bare independent events, then:

$P(A\space and\space B)=P(A)\times P(B)$

Events are independent if and only if 

$P(A \space and\space B)=P(A)×P(B).$

6.

For independence of two events A and B, we must have 

$P(A \space and\space B)=P(A).P(B)$

$P(A).P(B)=0.25\times0.60$

$=0.15$

$=P(A\space and\space B)$

Since P(A and B) = 0.15 = P(A).P(B), then the events A and B are independent.