Question

honors algebra ii
unit 1 test form a: probability
please show all of your work to receive full credit and write neatly. clearly indicate your final answer. thank you!
calculator use is allowed.
i can solve relevant real - world problems using probability.

1. a bag contains 5 red, 3 green, and 8 blue marbles. three are selected in sequence without replacement. what is the probability of selecting a red, a green, and a blue in that order? show how you found your answer.
2. how many 3 - digit patterns can be formed using the digits 5, 6, 7, 8, and 9, if the numbers can be used more than once? show how you found your answer.
3. how many 3 - digit patterns can be formed using the digits 5, 6, 7, 8, and 9, if the numbers can only be used once? show how you found your answer.
4. suppose you spin the spinner shown below three times. what is the probability of getting an even number all three times?
5. mrs. otte currently has 25 apps on her phone but needs to cut back to 18 to have sufficient storage. how many ways can mrs. otte select groups of 18 apps to keep on her phone? show how you found your answer.

Explanation:

Step1: Calculate probability for problem 1

The total number of marbles initially is \(5 + 3+8=16\). The probability of selecting a red first is \(\frac{5}{16}\), then for a green (with 15 marbles left) is \(\frac{3}{15}\), and for a blue (with 14 marbles left) is \(\frac{8}{14}\). The combined - probability is \(\frac{5}{16}\times\frac{3}{15}\times\frac{8}{14}=\frac{120}{3360}=\frac{1}{28}\).

Step2: Calculate number of 3 - digit patterns for problem 2

Since digits can be used more than once, for each of the three digit - positions, there are 5 choices. By the multiplication principle, the number of 3 - digit patterns is \(5\times5\times5 = 125\).

Step3: Calculate number of 3 - digit patterns for problem 3

Since digits can only be used once, we use the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 5\) and \(r=3\). \(P(5,3)=\frac{5!}{(5 - 3)!}=\frac{5!}{2!}=\frac{5\times4\times3\times2!}{2!}=60\).

Step4: Calculate probability for problem 4

The spinner has 8 sections, and 4 of them are even (\(2,4,6,8\)). The probability of getting an even number on one spin is \(\frac{4}{8}=\frac{1}{2}\). Since the spins are independent events, the probability of getting an even number three times is \(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\).

Step5: Calculate number of combinations for problem 5

We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 25\) and \(r = 18\). \(C(25,18)=C(25,7)=\frac{25!}{18!(25 - 18)!}=\frac{25!}{18!7!}=\frac{25\times24\times23\times22\times21\times20\times19}{7\times6\times5\times4\times3\times2\times1}=480700\).

Answer:

1. \(\frac{1}{28}\)
2. 125
3. 60
4. \(\frac{1}{8}\)
5. 480700