Question

the letters a, l, g, e, b, r, a are on 7 tiles in a bag, one letter on each tile. if you select tiles randomly from the bag and place them in a row from left to right, what is the probability the tiles will spell out algebra? (lesson 19.2) (1 point) a. $\frac{1}{1260}$ b. $\frac{1}{2520}$ c. $\frac{1}{21}$ d. $\frac{1}{7}$ 6. 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. $\frac{1}{715}$

Explanation:

Step1: Calculate the total number of permutations for the letters

The formula for permutations of \(n\) objects where \(a\) of one kind are identical is \(\frac{n!}{a!}\). Here \(n = 7\) (total number of tiles) and \(a=2\) (since there are 2 'A's). So the total number of permutations is \(\frac{7!}{2!}=\frac{7\times6\times5\times4\times3\times2\times1}{2\times1}=2520\).

Step2: Calculate the number of favorable permutations

There is only 1 way to spell "ALGEBRA" correctly.

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P = \frac{1}{2520}\).

for question 6:

Step1: Calculate the total number of combinations

The formula for combinations \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 13\) (total number of students) and \(r = 4\) (number of students to be chosen). So \(C(13,4)=\frac{13!}{4!(13 - 4)!}=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715\).

Step2: Calculate the number of favorable combinations

There is only 1 way to choose Ashley, Bob, Claire, and Daniel.

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P=\frac{1}{715}\).

Answer:

B. \(\frac{1}{2520}\)