Probability #2 - Questions and Answers

Question 1 The probability of event A is 0.4, the probability of (A∪B)ᶜ is 0.1, and the probability of the intersection of A and B is 0.2. What's the probability of event B? Answer Step 1 Given Data: The probability of event Aᶜ is: P(Aᶜ) = 0.4. Thus, P(A) = 1-P(Aᶜ) = 1-0.4 = 0.6 The probability of event (A∪B)ᶜ is: P((A∪B)ᶜ) = 0.1 The probability of event (A∩B) is: P(A∩B) = 0.2 The probability of event A is, P(A) = 1- P(Aᶜ) = 1 - 0.4 = 0.6 Step 2 The probability of event (A∪B) is, P(A∪B) = 1 - P((A∪B)ᶜ) = 1 - 0.1 = 0.9 The probability of event A union B is, P(A∪B) = P(A) + P(B) – P(A∩B) 0.9 = 0.6 + P(B) – 0.2 P(B) = 0.9 – 0.6 + 0.2 P(B) = 0.5 Thus, the probability of event B is 0.5. Question 2 Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat? Answer There are totally 5 seats in which Martha sitting in the middle (third) seat then there is only one possible chance to fill the middle seat. Now, for first seat there are four possible chance, second seat has three possible chance, fourth seat has 2 possible chance and fifth seat has one possible chance. The number of ways Martha sitting in the middle seat is, Number of ways = 4 × 3 × 1 × 2 × 1 = 24 Thus, there are 24 possible ways Martha sitting in the middle seat. Question 3 Uniform distribution is defined over the interval from 2 to 5. Find the probability of a value more than 2.6. Answer Probability of a value more than 2.6 If the value is more than 2.6, it means that the value is between 2.6 and 5. So, the base is 5- 2.6 = 2.4 P(2.6 < value < 5) = height × base = (1 / (b - a)) * (5-2.6) = (1 / (5 - 2)) * 2.4 = (1/3) * 2.4 = 2.4 / 3 = 0.8 Question 4 How many different ways can you make change for a quarter? (Different arrangements of the same coins are not counted separately.) Answer 1 There are three coins that are less than a quarter (0.25 ) i n v a l u e , t h a t i s a p e n n y ( 0.01 )invalue,thatisapenny(0.01 ), a nickel (0.05 ) a n d a d i m e ( 0.10 )andadime(0.10 ). Possibilities are then: 2 dimes 1 nickel, 2 dimes and 5 pennies, 1 dime and 3 nickels, 1 dime and 2 nickels and 5 pennies, 1 dime and 1 nickel and 10 pennies, 1 dime and 15 pennies, 5 nickels, 4 nickels and 5 pennies, 3 nickels and 10 pennies, 2 nickels and 15 pennies, 1 nickel and 20 pennies, 25 pennies Answer is 12 Answer 2 Explanation: If we use a quarter as the first coin, the remaining amount to be made is (25 - 25) = 0. In this case, there is only one way to make change: using a single quarter. If we use a dime as the first coin, the remaining amount to be made is (25 - 10) = 15. We can determine the number of ways to make change for 15, denoted as C(15), using the same method. If we use a nickel as the first coin, the remaining amount to be made is (25 - 5) = 20. We can determine the number of ways to make change for 20, denoted as C(20), using the same method. If we use a penny as the first coin, the remaining amount to be made is (25 - 1) = 24. We can determine the number of ways to make change for 24, denoted as C(24), using the same method. To find c(), we sum up the number of ways for each case: C() = C(0) + C(15) + C(20) + C(24) Using the same approach, we can break down each subproblem until we reach base cases that can be directly computed. The base cases are: C(0) = 1 (only one way to make no change) C(5) = 1 (only one way to make change with a nickel) C(10) = 1 (only one way to make change with a dime) C(15) = 2 (two ways: 1 dime and 1 nickel, or 3 nickels) C(20) = 2 (two ways: 2 dimes, or 1 dime and 2 nickels) C(24) = 4 (four ways: 2 dimes and 4 pennies, or 1 dime, 2 nickels, and 4 pennies, or 3 nickels and 9 pennies, or 1 dime and 14 pennies) Now, we can substitute these values back into the original equation: C()=1+2+2+4=9 Therefore, there are 9 different ways to make change for a quarter. Answer 3 We need to find the number of solutions for the equation 25q + 10d + 5n + p = 25, where q, d, n, p are non-negative integers. Using generating functions, we can express this equation as: (1 + x + x² + ...)(1 + x¹⁰ + x²⁰ + ...)(1 + x⁵ + x¹⁰ + ...)(1 + x) The coefficient of x²⁵ in this expression represents the number of ways to make change for a quarter. Simplifying the expression, we have: 1 / ( (1-x) (1-x¹⁰) (1 - x⁵) (1-x) ) Expanding this expression using partial fractions, we can write: A/(1 - x) + B/(1 - x¹⁰) + C/(1 - x⁵) + D/(1 - x) Solving for the values of A, B, C, and D, we can substitute them back into the expression and find the coefficient of x²⁵. Thus, the number of different ways to make change for a quarter is the coefficient of x²⁵ in the expanded expression. Question 5 How many bit strings of length seven either begin with two 0s or end with three 1s? Answer Let set A1 be the bit strings of length seven beginning with two zeros. Then |A1|= 1 * 1 * 2 * 2 * 2 * 2 * 2 = 2^5 = 32 Let set A2 be the bit strings of length seven ending with three ones. Then |A2| = 2 * 2 * 2 * 2 * 1 * 1 * 1 = 2^4 = 16 Let A1 ∩ A2 be the set of bit strings of length seven beginning with two zeros and ending with three ones. Then |A1 ∩ A2| = 1 * 1 * 2 * 2 * 1 * 1 * 1 = 2^2 = 4 |A1 ∪ A2| = |A1| + |A2| - |A1 ∩ A2| = 32 + 16 - 4 = 44