Probability and combinatorics #4 - Questions and Answers

Question 1 A coin will be tossed 100 times. You get to pick 11 numbers. If the number of heads turns out to equal one of your 11 numbers, you win a dollar. Which 11 numbers should you pick, and what is your chance (approximately) of winning? Explain. Answer Let 0 represent tails and let 1 represent heads. The 100 tosses are equivalent to drawing 100 tickets at random from a box that contains one ticket with the number 1 and one ticket with the number 0, since we have a 1 in 2 chance of getting heads. Numbers of draws=100 Expected value and standard deviation of box The total value of all the tickets divided by the total number of tickets gives us the expected value of the box, which is the first thing we calculate. Expected value box = (0+1)/2 = 1/2 = 0.5 Let us next determine the standard deviation using the shortcut formula: SD = (big number - small number) * √(fraction with big number * fraction with small number) = (1 - 0) * √(1/2 * 1/2) = 1 * √(1/4) = 1 * 1/2 = 0.5 Expected value and standard error of sum Next, the expected values for the sum is the product of the number of draws: = 100 * 0.5 = 50 The standard error of the sum is the product of the square root of the number of draws and the standard deviation of the box = √100 * 0.5 = 10 * 0.5 = 5 We note that our picked numbers are all within one standard error of the expected value. The chance of winning is the area corresponding to z=1.00 in the normal probability table in the appendix, which is thus 68.27% Question 2 Take a look at a group of 20 people. If everyone shakes hands with everyone else, how many handshakes take place? Answer 1 The two participants in a handshake are the only ones who define it. Therefore, among 20 people, there are as many handshakes as groups of two. The binomial is defined as the number of unorder groups of two among 20 different objects. (20 choose 2) = 20!/(2!(20-2)!) = (20*19)/2 = 190 Answer 2 Step 1: To find the number of handshakes that take place in a group of 20 people, we can use the formula for the sum of the first n natural numbers. In this case, n = 19, since one person cannot shake hands with themselves. The formula for the sum of the first n natural numbers is given by: ∑(k from 1 to n) = n(n+1) / 2 Step 2: Substituting n = 19 into the formula, we get: ∑(k from 1 to 19) = 19(19+1)/2 = (19*20)/2 = 190 Therefore, there would be 190 handshakes taking place among 20 people. Answer 3 Explanation: Let's consider the first person in the group. This person can shake hands with 19 other people in the group (excluding themselves). The second person can shake hands with the remaining 18 people (excluding themselves and the person they already shook hands with). Similarly, the third person can shake hands with the remaining 17 people, and so on. We can see that the number of handshakes decreases by one for each person in the group. Therefore, the total number of handshakes H can be calculated using the formula: H = 19 + 18 + 17 + ... + 2 + 1 We can rewrite this formula using summation notation as: H = ∑(n from 1 to 19) n Now, let's calculate the value of H: H = ∑(n from 1 to 19) n = 19*(19+1)/2 = 19*20/2 = 190 Thus, there will be 190 handshakes taking place among the group of 20 people. Answer 4 To solve the problem, we can use a combination formula to calculate the number of handshakes. The combination formula is given by: C(n,r) = n! / (r!(n-r)!) where C(n,r) represents the number of combinations of r items chosen from a set of n items, and n! denotes the factorial of n. In this case, we have a group of 20 people, and we want to find the number of handshakes. Each person shakes hands with every other person, so we need to choose 2 people at a time from the group of 20. Using the combination formula, we have: C(20,2) = 20! / (2!(20-2)!) Simplifying further: C(20,2) = 20! / (2!18!) Since 2! = 2, we can substitute this value: C(20,2) = 20! / (218!) Now, let's calculate the factorial: C(20,2) = (201918!) / (218!) Simplifying the expression: C(20,2) = (2019)/2 C(20,2) = 190 Therefore, there are 190 handshakes that take place among the 20 people in the group. Question 3 You arrive at a bus stop at 10 o'clock, knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30. (a) What is the probability that you will have to wait longer than 10 minutes? (b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes? Answer Given that Bus arrives uniformly at the bus stop between 10am and 10:30am Let X denote the arrival of bus at the bus stop. Then, X is a uniform random variable having density as f(x) = 1/30, x ∈ [0,30] a) Probability that a passenger arriving at the bus stop between 10 and 10:30 will have to wait more than 10 minutes. P[X>10] = ∫(from 10 to 30) f(x)dx = ∫(from 10 to 30) 1/30 dx = 1/30 * [x](from 10 to 30) = 1/30 * (30 - 10) = 20/30 = 2/3 b) At 10:15am, bus hasn’t Question 4 Suppose that f(x) = x/8 for 3 < x < 5. Determine the following probabilities: a. P(X < 4) b. P(X > 3.5) c. P(4 < X < 5) d. P(X < 4.5) e. P(X < 3.5 or X > 4.5) Answer 1 The probability density function f of the random variable X is given as: f(x) = x/8, 3 < x < 5 Calculate the general probability P(a 4.5) = P(3) Answer 2 Given that f(x) = x/8 for 3 < x < 5, we can calculate the CDF as follows: F(x) = ∫(from 3 to x) f(t)dt For 3 < x < 5, the function f(x) is a linear function, so we can easily integrate it within that range: F(x) = ∫(from 3 to x) t/8 dt Simplifying the integral: F(x) = (1/8) * [t²/2](from 3 to x) F(x) = (1/8) * (x²/2 - 9/2) F(x) = (x²/16) - (9/16) F(x) = (x² - 9)/16 Now, let's calculate the probabilities using the CDF we found. a. P(X < 4) To find this probability, we substitute x = 4 into the CDF: P(X < 4) = F(4) = (4² - 9)/16 Simplifying the expression: P(X < 4) = (16 - 9)/16 P(X < 4) = 7/16 Therefore, P(X < 4) = 7/16 b. P(X > 3.5) To find this probability, we subtract the probability of X being less than or equal to 3.5 from 1 (since the total probability must add up to 1): P(X > 3.5) = 1 - P(X ≤ 3.5) Using the CDF, we can calculate: P(X > 3.5) = 1 - F(3.5) P(X > 3.5) = 1 - ((3.5² - 9)/16) Simplifying the expression: P(X > 3.5) = 1 - ((12.25 - 9)/16) P(X > 3.5) = 1 - (3.25/16) P(X > 3.5) = 1 - (13/64) P(X > 3.5) = 51/64 c. P(4 < X < 5) To find this probability, we subtract the probability of X being less than or equal to 4 from the probability of X being less than or equal to 5: P(4 < X < 5) = F(5) - F(4) P(4 < X < 5) = ((5²-9)/16) - ((4²-9)/16) Simplifying the expression: P(4 < X < 5) = ((25-9)/16) - ((16-9)/16) P(4 < X < 5) = (16/16) - (7/16) P(4 < X < 5) = 9/16 Therefore, P(4 < X < 5) = 9/16 d. P(X < 4.5) To find this probability, we substitute x = 4.5 into the CDF: P(X < 4.5) = F(4.5) = (4.5² - 9)/16 Simplifying the expression: P(X < 4.5) = (20.25 - 9)/16 P(X < 4.5) = 11.25/16 P(X < 4.5) = 11.25/16 P(X < 4.5) = 11.25/16 P(X < 4.5) = 5.0625/8 P(X < 4.5) = 81/128 Therefore, P(X < 4.5) = 81/128 e. P(X < 3.5 or X > 4.5) To find this probability, we calculate the sum of the probabilities P(X < 3.5) and P(X > 4.5). Since these two events are mutually exclusive, we can add their probabilities: P(X < 3.5 or X > 4.5) = P(X < 3.5) + P(X > 4.5) P(X < 3.5 or X > 4.5) = F(3.5) + (1-F(4.5)) Substituting the CDF values: P(X < 3.5 or X > 4.5) = ((3.5² - 9)/16) + (1 - ((4.5² - 9)/16)) Simplifying the expression: P(X < 3.5 or X > 4.5) = (12.25 - 9)/16 + (1 - (20.25 - 9)/16) P(X < 3.5 or X > 4.5) = (3.25/16) + (1 - (11.25/16)) P(X < 3.5 or X > 4.5) = (13/64) + (1 - 81/128) P(X < 3.5 or X > 4.5) = (13/64) + (128/128 - 81/128) P(X < 3.5 or X > 4.5) = 13/64 + 47/128 P(X < 3.5 or X > 4.5) = (26/128) + 47/128 P(X < 3.5 or X > 4.5) = 73/128 Answer 3 a. To find P(X < 4), we need to calculate the integral of f(x) from -∞ to 4: P(X < 4) = ∫(from 3 to 4) f(x) dx = ∫(from 3 to 4) x/8 dx. b. To find P(X > 3.5), we need to calculate the integral of f(x) from 3.5 to ∞ : P(X > 3.5) = ∫(from 3.5 to 5) f(x) dx = ∫(from 3.5 to 5) x/8 dx. c. To find P(4 < X < 5), we need to calculate the integral of f(x) from 4 to 5: P(4 < X < 5) = ∫(from 4 to 5) f(x)dx = ∫(from 4 to 5) x/8 dx. d. To find P(X < 4.5), we need to calculate the integral of f(x) from -∞ to 4.5: P(X < 4.5) = ∫(from 3 to 4.5) f(x)dx = ∫(from 3 to 4.5) x/8 dx. e. To find P(X < 3.5 or X > 4.5), we need to calculate the sum of the probabilities of P(X < 3.5) and P(X > 4.5): P(X < 3.5 or X > 4.5) = P(X < 3.5) + P(X > 4.5). Answer 4 Step 1: a. P(X < 4): To find the probability that x is less than 4, we need to calculate the definite integral of f(x) over the interval (3, 4). The probability can be obtained as follows: P(X < 4) = ∫(from 3 to 4) f(x)dx = ∫(from 3 to 4) x/8 dx Evaluating the integral: P(X < 4) = [x²/16](from 3 to 4) = (4²/16) - (3²/16) = (16-9)/16 = 7/16 Therefore, P(X < 4) = 7/16 Step 2: b. P(X > 3.5): To find the probability that X is greater than 3.5, we need to calculate the definite integral of f(x) over the interval (3.5, 5). The probability can be obtained as follows: P(X > 3.5) = ∫(from 3.5 to 5) f(x) dx = ∫(from 3.5 to 5) x/8 dx Evaluating the integral: P(X > 3.5) = [x²/16](from 3.5 to 5) = (5²/16) - (3.5²/16) = (25-12.25)/16 = 12.75/16 = 51/64 Therefore, P(X > 3.5) = 51/64 Step 3: c. P(4 < X < 5): To find the probability that X lies between 4 and 5, we need to calculate the definite integral of f(x) over the interval (4, 5). The probability can be obtained as follows: P(4 < X < 5) = ∫(from 4 to 5) f(x) dx = ∫(from 4 to 5) x/8 dx Evaluating the integral: P(4 < X < 5) = [x²/16](from 4 to 5) = (5²/16) - (4²/16) = (25-16)/16 = 9/16 Therefore, P(4 < X < 5) = 9/16 Step 4: d. P(X < 4.5): To find the probability that X is less than 4.5, we need to calculate the definite integral of f(x) over the interval (3, 4.5). The probability can be obtained as follows: P(X < 4.5) = ∫(from 3 to 4.5) f(x) dx = ∫(from 3 to 4.5) x/8 dx Evaluating the integral: P(X < 4.5) = [x²/16](from 3 to 4.5) = (4.5²/16) - (3²/16) = (20.25-9)/16 = 11.25/16 = 9/12 Therefore, P(X < 4.5) = 9/16 Step 5: e. P(X < 3.5 or X > 4.5): To find the probability that X is less than 3.5 or greater than 4.5, we can calculate the sum of the probabilities P(X < 3.5) and P(X > 4.5). These probabilities can be obtained using the same approach as in parts (a) and (b): P(X < 3.5) = ∫(from 3 to 3.5) f(x) dx = ∫(from 3 to 3.5) x/8 dx P(X > 4.5) = ∫(from 4.5 to 5) f(x)dx = ∫(from 4.5 to 5) x/8 dx Evaluating the integrals: P(X < 3.5) = [x²/16](from 3 to 3.5) = (3.5²/16) - (3²/16) = (12.25 - 9)/16 = 3.25/16 P(X > 4.5) = [x²/16](from 4.5 to 5) = (5²/16) - (4.5²/16) = (25 - 20.25)/16 = 4.75/16 Therefore, P(X < 3.5 or X > 4.5) = P(X < 3.5) + P(X > 4.5) = 3.25/16 + 4.75/16 = 8/16 = 1/2