QUESTION IMAGE
Question
1 numeric 10 points
there are 12 books on a summer reading list and you are required to read 3 of them. how many choices of 3 books are possible?
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2 numeric 10 points
amy has just opened a new store. if successful, she can expect an annual income of $100,000. if unsuccessful, she will lose $175,000. if the probability of success is 0.82, find her expected income for the first year.
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3 numeric 10 points
if there are 11 people on a baseball team, determine how many different ways a 1st baseman, a 2nd baseman and a 3rd baseman could be chosen.
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Question 1
Step1: Identify the problem type
This is a combination problem since the order of choosing the books does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 12 \) (total number of books) and \( k=3 \) (number of books to choose).
Step2: Calculate the factorial values
First, calculate \( n!=12! = 12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \), \( k!=3! = 3\times2\times1 \), and \( (n - k)!=(12 - 3)!=9! = 9\times8\times7\times6\times5\times4\times3\times2\times1 \).
Step3: Substitute into the combination formula
\( C(12, 3)=\frac{12!}{3!(12 - 3)!}=\frac{12\times11\times10\times9!}{3\times2\times1\times9!} \). The \( 9! \) terms cancel out. Then \( \frac{12\times11\times10}{3\times2\times1}=\frac{1320}{6} = 220 \).
Step1: Recall the expected value formula
The expected value \( E(X) \) for a discrete random variable is given by \( E(X)=x_1P(x_1)+x_2P(x_2) \), where \( x_1 \) and \( x_2 \) are the possible outcomes and \( P(x_1) \), \( P(x_2) \) are their respective probabilities. Here, \( x_1 = 100000 \) (income if successful), \( P(x_1)=0.82 \), \( x_2=- 175000 \) (loss if unsuccessful), and \( P(x_2)=1 - 0.82 = 0.18 \).
Step2: Calculate the expected value
\( E(X)=(100000\times0.82)+(- 175000\times0.18) \). First, calculate \( 100000\times0.82 = 82000 \) and \( - 175000\times0.18=-31500 \). Then add these two results: \( 82000-31500 = 50500 \).
Step1: Identify the problem type
This is a permutation problem since the order of choosing the basemen (1st, 2nd, 3rd) matters. The formula for permutations is \( P(n, k)=\frac{n!}{(n - k)!} \), where \( n = 11 \) (total number of people) and \( k = 3 \) (number of basemen to choose).
Step2: Calculate the factorial values
\( n!=11! = 11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \), \( (n - k)!=(11 - 3)!=8! = 8\times7\times6\times5\times4\times3\times2\times1 \).
Step3: Substitute into the permutation formula
\( P(11, 3)=\frac{11!}{(11 - 3)!}=\frac{11\times10\times9\times8!}{8!} \). The \( 8! \) terms cancel out. So \( 11\times10\times9=990 \).
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