Question

1. a bag contains 21 red marbles and 4 blue marbles. two marbles will be randomly selected from the bag. the first marble will not be returned to the bag before the second marble is selected.

a. given that the first marble selected is blue, what is the probability that the second marble selected will also be blue?
b. what is the probability that someone selects two red marbles?

1. in a class of 28 students, 10 have a brother and 12 have a sister. there are 10 students who do not have any siblings. what is the probability that a student has a sister given that they do not have a brother?

Explanation:

Problem 2a

Step1: Identify the situation

Since the first marble is not returned, it's a dependent event. The first marble is blue, so we need to find the probability the second is blue given the first was blue.

Step2: Calculate the new total and blue marbles

Original total marbles: \(21 + 4 = 25\). After removing one blue marble, total marbles left: \(25 - 1 = 24\). Blue marbles left: \(4 - 1 = 3\).

Step3: Find the probability

Probability = \(\frac{\text{Blue marbles left}}{\text{Total marbles left}} = \frac{3}{24} = \frac{1}{8}\).

Step1: Determine the type of event

We are selecting two red marbles without replacement (since first is not returned), so it's a dependent event.

Step2: Calculate the probability of first red

Probability of first red: \(\frac{21}{25}\) (21 red out of 25 total).

Step3: Calculate the probability of second red

After removing one red marble, red marbles left: \(21 - 1 = 20\), total marbles left: \(25 - 1 = 24\). Probability of second red: \(\frac{20}{24}\).

Step4: Multiply the probabilities

Total probability = \(\frac{21}{25} \times \frac{20}{24}\). Simplify: \(\frac{21 \times 20}{25 \times 24} = \frac{420}{600} = \frac{7}{10}\).

Step1: Identify the given values

Total students: 28. Students with no siblings: 10. Students with a sister: 12 (but we need those with sister and no siblings? Wait, no: Wait, the problem says "10 have a brother and 12 have a sister. There are 10 students who do not have any siblings." Wait, maybe re - read: Wait, the problem is "What is the probability that a student has a sister given that they do not have a brother?" Wait, no, original problem: "In a class of 28 students, 10 have a brother and 12 have a sister. There are 10 students who do not have any siblings. What is the probability that a student has a sister given that they do not have a brother?"

Wait, let's re - analyze:

First, find the number of students who do not have a brother. Total students: 28. Students with a brother: 10. So students without a brother: \(28 - 10 = 18\).

Now, among these 18 (without a brother), how many have a sister? Wait, students with no siblings: 10. So students without a brother but with siblings: \(18 - 10 = 8\)? Wait, no, maybe the 12 with a sister: Wait, maybe the correct way:

Wait, the problem is "given that they do not have a brother", so the sample space is students without a brother.

Total students: 28. Students with a brother: 10, so students without a brother: \(28 - 10 = 18\).

Students with no siblings: 10. So students without a brother and with siblings: \(18 - 10 = 8\)? Wait, no, the 12 have a sister: Wait, maybe the 12 with a sister include those with or without a brother. Wait, maybe the number of students without a brother and with a sister:

Total students with a sister: 12. Students with a sister and a brother: Let's see, total with siblings: \(28 - 10 = 18\). Students with a brother: 10, students with a sister: 12. Using inclusion - exclusion: students with at least one sibling (brother or sister) = students with brother + students with sister - students with both. We know students with at least one sibling: \(28 - 10 = 18\). So \(18 = 10 + 12 - \text{both}\), so \(\text{both} = 10 + 12 - 18 = 4\). So students with a sister only (no brother): \(12 - 4 = 8\).

So the number of students without a brother and with a sister is 8, and the number of students without a brother is 18.

Step2: Calculate the probability

Probability = \(\frac{\text{Students without a brother and with a sister}}{\text{Students without a brother}}=\frac{8}{18}=\frac{4}{9}\).

Wait, maybe I made a mistake. Let's re - do:

Wait, the problem says "10 have a brother and 12 have a sister. There are 10 students who do not have any siblings."

So total students: 28.

Students with no siblings: 10. So students with siblings: \(28 - 10 = 18\).

Students with a brother: 10, students with a sister: 12.

Let \(B\) be the set of students with a brother, \(S\) be the set with a sister.

\(n(B\cup S)=n(B)+n(S)-n(B\cap S)\)

We know \(n(B\cup S) = 18\) (students with siblings), \(n(B) = 10\), \(n(S)=12\).

So \(18 = 10 + 12 - n(B\cap S)\)

\(n(B\cap S)=10 + 12 - 18 = 4\)

So students with a sister but no brother: \(n(S)-n(B\cap S)=12 - 4 = 8\)

Students without a brother: \(28 - n(B)=28 - 10 = 18\)

So the probability that a student has a sister given they do not have a brother is \(\frac{8}{18}=\frac{4}{9}\)

Step1: Define the conditional probability

We need \(P(S|\text{not }B)\), where \(S\) is having a sister and \(\text{not }B\) is not having a brother.

Step2: Find \(n(\text{not }B)\) and \(n(S\cap\text{not }B)\)

\(n(\text{not }B)=28 - 10 = 18\)

\(n(S\cap\text{not }B)=n(S)-n(S\cap B)=12 - 4 = 8\) (since \(n(S\cap B) = 4\) as calculated above)

Ste…

Answer:

\(\frac{1}{8}\)

Problem 2b