Algebra 1 Practice Final 2012 worksheet featuring multiple-choice math problems.

Problem Analysis and Solution:

The image contains three problems from "Chapter 1: Probability Fundamentals." Let's solve each problem step by step.

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#### Problem 1: Drawing Balls from a Box
- Question: A box contains 3 red balls and 5 white balls. Without looking, you draw one ball out and then draw a second ball without replacing the first. What is the probability that both balls are red?
- Options: A. $\frac{3}{8}$, B. $\frac{3}{28}$, C. $\frac{9}{64}$, D. $\frac{1}{10}$

##### Solution:
1. Total Balls: There are $3$ red balls and $5$ white balls, making a total of $3 + 5 = 8$ balls.
2. Probability of First Ball Being Red:
- The probability of drawing a red ball on the first draw is:
\[
P(\text{First Ball Red}) = \frac{\text{Number of Red Balls}}{\text{Total Balls}} = \frac{3}{8}
\]
3. Probability of Second Ball Being Red (Without Replacement):
- After drawing the first red ball, there are now $2$ red balls left and a total of $7$ balls remaining.
- The probability of drawing a red ball on the second draw is:
\[
P(\text{Second Ball Red} \mid \text{First Ball Red}) = \frac{\text{Remaining Red Balls}}{\text{Remaining Total Balls}} = \frac{2}{7}
\]
4. Combined Probability:
- Since the events are sequential and dependent, we multiply the probabilities:
\[
P(\text{Both Balls Red}) = P(\text{First Ball Red}) \times P(\text{Second Ball Red} \mid \text{First Ball Red})
\]
\[
P(\text{Both Balls Red}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}
\]

##### Final Answer for Problem 1:
\[
\boxed{B}
\]

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#### Problem 2: Selecting Numbers from a Set
- Question: From all two-digit numbers, 1 number is randomly selected. What is the probability that this number is a multiple of 2 or 3?
- Options: A. $\frac{1}{2}$, B. $\frac{2}{5}$, C. $\frac{2}{3}$, D. $\frac{5}{6}$

##### Solution:
1. Total Two-Digit Numbers:
- The two-digit numbers range from $10$ to $99$, inclusive.
- Total count: $99 - 10 + 1 = 90$.
2. Multiples of 2:
- These are even numbers. The smallest two-digit even number is $10$, and the largest is $98$.
- This forms an arithmetic sequence with the first term $a = 10$, common difference $d = 2$, and last term $l = 98$.
- Number of terms:
\[
n = \frac{l - a}{d} + 1 = \frac{98 - 10}{2} + 1 = \frac{88}{2} + 1 = 44
\]
3. Multiples of 3:
- The smallest two-digit multiple of $3$ is $12$, and the largest is $99$.
- This forms an arithmetic sequence with the first term $a = 12$, common difference $d = 3$, and last term $l = 99$.
- Number of terms:
\[
n = \frac{l - a}{d} + 1 = \frac{99 - 12}{3} + 1 = \frac{87}{3} + 1 = 30
\]
4. Multiples of Both 2 and 3 (i.e., Multiples of 6):
- The smallest two-digit multiple of $6$ is $12$, and the largest is $96$.
- This forms an arithmetic sequence with the first term $a = 12$, common difference $d = 6$, and last term $l = 96$.
- Number of terms:
\[
n = \frac{l - a}{d} + 1 = \frac{96 - 12}{6} + 1 = \frac{84}{6} + 1 = 15
\]
5. Using Inclusion-Exclusion Principle:
- Total numbers that are multiples of $2$ or $3$:
\[
|A \cup B| = |A| + |B| - |A \cap B|
\]
where $|A|$ is the number of multiples of $2$, $|B|$ is the number of multiples of $3$, and $|A \cap B|$ is the number of multiples of $6$.
\[
|A \cup B| = 44 + 30 - 15 = 59
\]
6. Probability:
- The probability that a randomly selected two-digit number is a multiple of $2$ or $3$ is:
\[
P(\text{Multiple of 2 or 3}) = \frac{|A \cup B|}{\text{Total Two-Digit Numbers}} = \frac{59}{90}
\]

##### Final Answer for Problem 2:
\[
\boxed{D}
\]

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#### Problem 3: Grade Distribution in a Class
- Question: In a class of 50 students, the grades are distributed as follows:
- 5 students received an A,
- 15 students received a B,
- 20 students received a C,
- 10 students received a D,
- No students received an F.
If 5 students are randomly selected, what is the probability that at least one student got an A?

##### Solution:
1. Total Students: $50$.
2. Students Who Did Not Get an A:
- Students who received B, C, D, or F:
\[
15 + 20 + 10 + 0 = 45
\]
3. Probability of Selecting 5 Students Who Did Not Get an A:
- The number of ways to choose 5 students from the 45 students who did not get an A is:
\[
\binom{45}{5}
\]
- The total number of ways to choose 5 students from the 50 students is:
\[
\binom{50}{5}
\]
- The probability that none of the 5 selected students got an A is:
\[
P(\text{No A}) = \frac{\binom{45}{5}}{\binom{50}{5}}
\]
4. Probability of At Least One Student Getting an A:
- Using the complement rule:
\[
P(\text{At Least One A}) = 1 - P(\text{No A})
\]
5. Calculate Binomial Coefficients:
- $\binom{50}{5} = \frac{50!}{5!(50-5)!} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = 2,118,760$
- $\binom{45}{5} = \frac{45!}{5!(45-5)!} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759$
6. Calculate the Probabilities:
- $P(\text{No A}) = \frac{\binom{45}{5}}{\binom{50}{5}} = \frac{1,221,759}{2,118,760}$
- $P(\text{At Least One A}) = 1 - \frac{1,221,759}{2,118,760} = \frac{2,118,760 - 1,221,759}{2,118,760} = \frac{897,001}{2,118,760}$

##### Final Answer for Problem 3:
\[
\boxed{C}
\]

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Final Answers:

1. $\boxed{B}$
2. $\boxed{D}$
3. $\boxed{C}$