Question

1-87c. use the combinations you found earlier to find
the following probabilities.

note: write your probability as a fraction.

hint: \\(\frac{\text{number of successes}}{\text{total number of combinations}}\\)

a. find the probability that renae will listen to two
songs with the name “mama” in the title.

the probability is \\(\square\\).

b. what is the probability that at least one of the
songs will have the name “mama” in the title?

the probability is \\(\square\\).

Explanation:

To solve these probability problems, we need to know the number of songs with "Mama" in the title (let's call this \( n \)) and the total number of songs (let's call this \( N \)), and the number of songs we are choosing (let's assume it's 2, as it's about two songs). However, since the "combinations found earlier" are not provided, we'll assume a common scenario (e.g., 2 songs with "Mama" and 4 total songs, choosing 2).

Part (a)

Step 1: Define Variables

Let’s assume:

• Number of "Mama" songs (\( k \)) = 2
• Total songs (\( N \)) = 4
• Songs chosen (\( r \)) = 2

Step 2: Calculate Successful Combinations

We need to choose 2 "Mama" songs from 2:
\[
\binom{k}{r} = \binom{2}{2} = \frac{2!}{2!(2-2)!} = 1
\]

Step 3: Calculate Total Combinations

Total ways to choose 2 songs from 4:
\[
\binom{N}{r} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

Step 4: Calculate Probability

Probability = \(\frac{\text{Successful Combinations}}{\text{Total Combinations}} = \frac{1}{6}\)

Part (b)

Step 1: Define "At Least One"

"At least one" = 1 or 2 "Mama" songs. We can calculate it as \( 1 - P(\text{no "Mama" songs}) \).

Step 2: Calculate Combinations for No "Mama" Songs

Number of non-"Mama" songs = \( 4 - 2 = 2 \). Choose 2 from 2:
\[
\binom{2}{2} = 1
\]

Step 3: Calculate Probability of No "Mama" Songs

\[
P(\text{no "Mama"}) = \frac{\binom{2}{2}}{\binom{4}{2}} = \frac{1}{6}
\]

Step 4: Calculate Probability of At Least One

\[
P(\text{at least one}) = 1 - \frac{1}{6} = \frac{5}{6}
\]

Final Answers

a. \(\boldsymbol{\frac{1}{6}}\) (assuming 2 "Mama" songs, 4 total, choosing 2)
b. \(\boldsymbol{\frac{5}{6}}\) (same assumptions)

Note: If the actual number of songs differs, substitute \( k \), \( N \), and \( r \) with the correct values from the "earlier combinations."

Answer:

To solve these probability problems, we need to know the number of songs with "Mama" in the title (let's call this \( n \)) and the total number of songs (let's call this \( N \)), and the number of songs we are choosing (let's assume it's 2, as it's about two songs). However, since the "combinations found earlier" are not provided, we'll assume a common scenario (e.g., 2 songs with "Mama" and 4 total songs, choosing 2).

Part (a)

Step 1: Define Variables

Let’s assume:

• Number of "Mama" songs (\( k \)) = 2
• Total songs (\( N \)) = 4
• Songs chosen (\( r \)) = 2

Step 2: Calculate Successful Combinations

We need to choose 2 "Mama" songs from 2:
\[
\binom{k}{r} = \binom{2}{2} = \frac{2!}{2!(2-2)!} = 1
\]

Step 3: Calculate Total Combinations

Total ways to choose 2 songs from 4:
\[
\binom{N}{r} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

Step 4: Calculate Probability

Probability = \(\frac{\text{Successful Combinations}}{\text{Total Combinations}} = \frac{1}{6}\)

Part (b)

Step 1: Define "At Least One"

"At least one" = 1 or 2 "Mama" songs. We can calculate it as \( 1 - P(\text{no "Mama" songs}) \).

Step 2: Calculate Combinations for No "Mama" Songs

Number of non-"Mama" songs = \( 4 - 2 = 2 \). Choose 2 from 2:
\[
\binom{2}{2} = 1
\]

Step 3: Calculate Probability of No "Mama" Songs

\[
P(\text{no "Mama"}) = \frac{\binom{2}{2}}{\binom{4}{2}} = \frac{1}{6}
\]

Step 4: Calculate Probability of At Least One

\[
P(\text{at least one}) = 1 - \frac{1}{6} = \frac{5}{6}
\]

Final Answers

a. \(\boldsymbol{\frac{1}{6}}\) (assuming 2 "Mama" songs, 4 total, choosing 2)
b. \(\boldsymbol{\frac{5}{6}}\) (same assumptions)

Note: If the actual number of songs differs, substitute \( k \), \( N \), and \( r \) with the correct values from the "earlier combinations."